## Electric Lines of Force-

Before you go through this article, make sure that you have gone through the previous article on Electric Lines of Force.

We have discussed-

• Electric lines of force represent the electric field around the charge.
• Electric lines of force are the imaginary curves.
• But the field they represent is real.

## Electric Flux-

 The total number of electric lines of force passing perpendicularly through a given area held inside an electric field is called as electric flux.

Electric flux is denoted by the symbol ΦE.

The electric flux passing through any given area depends upon the following factors-

Also Read- Electric Lines of Force

### 1. Electric Field Strength-

Electric flux ∝ Strength of Electric Field

This is because-

• Stronger the electric field, closer are the electric lines of force.
• Closer the electric lines of force, greater the amount of electric flux passing through the surface.

### 2. Area of The Surface-

Electric flux ∝ Area of the Surface

This is because-

• Greater the area of the surface, greater the amount of electric lines of force passing through it.
• Therefor, greater is the electric flux passing through the area.

### 3. Angle between Electric Field & Area Vector-

Electric flux ∝ cosθ

This is because-

• As the angle (θ) between electric field and area vector increases, the number of electric lines of force passing through the area decreases.

On combining these factors, we get the following electric flux formula.

## Electric Flux Formula-

Electric flux ΦE passing through any given area $\overrightarrow{A}$ held inside an electric field of strength $\overrightarrow{E}$ is given by-

• The SI unit of electric flux is NC-1m2.
• Electric flux is a scalar quantity.

## Electric Flux Density-

Electric flux passing per unit area through the surface is called as electric flux density.

Mathematically,

• The SI unit of electric flux density is NC-1.
• Electric flux density is a scalar quantity.

## Important Notes-

It is worth mentioning the following notes-

### Note-01:

• Area vector for a 2D surface is always perpendicular to its surface.
• Area vector for a 3D surface is always outward normal to its surface.

### Note-02:

• The formula ΦE = $\overrightarrow{E}.\overrightarrow{A}$ = EAcosθ is valid if and only if the value of electric field E at every point lying on the area A is same.

### Note-03:

• The formula ΦE = ∫$\overrightarrow{E}.\overrightarrow{dA}$ is valid everywhere.
• When electric field is uniform, $\overrightarrow{E}$ comes out of the integration and the formula reduces to ΦE = $\overrightarrow{E}.\overrightarrow{A}$.

## Problem-01:

Calculate the electric flux passing through the surface kept in the electric field with its plane perpendicular to the electric field as shown-

## Solution-

We have-

Given-

• Electric field strength (E) = 100 N/C
• Area of the surface (A) = 2 m x 2 m = 4 m2
• Angle between $\overrightarrow{E}$ and $\overrightarrow{A}$ = 0°

We know, electric flux ΦE passing through any surface is given by-

ΦE = EAcosθ

Substituting the values, we get-

ΦE = 100 x 4 x cos 0°

ΦE = 100 x 4 x 1

ΦE = 400 NC-1m2

Thus, electric flux passing through the given surface = 400 NC-1m2.

## Problem-02:

Calculate the electric flux passing through the surface kept in the electric field with its plane parallel to the electric field as shown-

## Solution-

We have-

Given-

• Electric field strength (E) = 100 N/C
• Area of the surface (A) = 2 m x 2 m = 4 m2
• Angle between $\overrightarrow{E}$ and $\overrightarrow{A}$ = 90°

We know, electric flux ΦE passing through any surface is given by-

ΦE = EAcosθ

Substituting the values, we get-

ΦE = 100 x 4 x cos 90°

ΦE = 100 x 4 x 0

ΦE = 0

Thus, no electric flux passes through the given surface.

## Problem-03:

Calculate the electric flux passing through the circular disk of radius 7 m placed in the electric field as shown-

## Solution-

We have-

Given-

• Electric field strength (E) = 100 N/C
• Radius of circular disk (R) = 7 m
• Angle between $\overrightarrow{E}$ and $\overrightarrow{A}$ = 60°

Now,

Area of the disk surface (A)

= πR2

= (22/7) x 7 x 7

= 154 m2

We know, electric flux ΦE passing through any surface is given by-

ΦE = EAcosθ

Substituting the values, we get-

ΦE = 100 x 154 x cos 60°

ΦE = 100 x 154 x 0.5

ΦE = 7700 NC-1m2

Thus, electric flux passing through the given circular disk = 7700 NC-1m2.

## Problem-04:

If $\overrightarrow{E}$ = $6\hat{i}&space;+&space;3\hat{j}&space;+&space;4\hat{k}$, calculate the electric flux through a surface of area 20 units in Y-Z plane.

## Solution-

Given-

• Electric field ($\overrightarrow{E}$) = $6\hat{i}&space;+&space;3\hat{j}&space;+&space;4\hat{k}$
• Area of the surface ($\overrightarrow{A}$) = $20\hat{i}$

We know, electric flux ΦE passing through any surface is given by-

ΦE = $\overrightarrow{E}.\overrightarrow{A}$

Substituting the values, we get-

ΦE = ($6\hat{i}&space;+&space;3\hat{j}&space;+&space;4\hat{k}$).$20\hat{i}$

ΦE = 120 units

Thus, electric flux passing through the given surface = 120 units.

## Problem-05:

A cylinder of radius R is placed in a uniform electric field E with its axis parallel to the field as shown-

Calculate the electric flux passing through-

1. Left plane surface
2. Right plane surface
3. Curved surface
4. Complete cylinder

## Solution-

Given-

• Electric field strength = E
• Radius of the cylinder = R

We know, electric flux ΦE passing through any surface is given by-

ΦE = EAcosθ

### Part-01: Electric Flux Through Left Plane Surface-

Electric flux through left plane surface (Φ1)

= E x πR2 x cos 180°

= -EπR2

Thus, electric flux passing through the left plane surface = -EπR2.

### Part-02: Electric Flux Through Right Plane Surface-

Electric flux through right plane surface (Φ2)

= E x πR2 x cos 0°

= EπR2

Thus, electric flux passing through the right plane surface = EπR2.

### Part-03: Electric Flux Through Curved Surface-

Electric flux through curved surface (Φ3)

= E x 2πRL x cos 90°

= 0

Thus, electric flux passing through the curved surface = 0.

### Part-04: Electric Flux Through Complete Cylinder-

Electric flux through complete cylinder (ΦE)

= Φ1 + Φ2 + Φ3

= -EπR2 + EπR2 + 0

= 0

Thus, electric flux passing through the complete cylinder = 0.

## Problem-06:

A positive charge of 17.7 μC is placed at the center of a hollow sphere of radius 0.5 m. Calculate the flux density through the surface of the sphere.

## Solution-

Given-

• Radius of hollow sphere (R) = 0.5 m
• Charge at center (Q) = 17.7 μC

### Step-01: Calculating Electric Field At Surface-

Electric field at the surface of a hollow sphere due to the charge kept at the center is constant.

The electric field at the surface is directed outward normal.

It is given by-

Substituting the values, we get-

Electric field at surface

= (9 x 109 x 17.7 x 10-6) / (0.5)2

= 637.2 x 103

= 6.37 x 105 N/C

### Step-02: Calculating Electric Flux Through Surface-

We have-

Electric flux passing through surface of the sphere

= EAcosθ

= E x 4πR2 x cosθ

= 6.37 x 105 x (4 x 3.14 x 0.5 x 0.5) x cos0°

= 20 x 105 NC-1m2

### Step-03: Calculating Electric Flux Density-

Electric flux density

= Total electric flux passing through the surface / Area of the surface

= 20 x 105 / (4 x 3.14 x 0.5 x 0.5)

= 20 x 105 / 3.14

= 6.36 x 105

= 6.4 x 105 N/C

Thus, electric flux density through the surface = 6.4 x 105 N/C.

To gain better understanding about Electric Flux,

Watch this Video Lecture

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## Electric Field-

Before you go through this article, make sure that you have gone through the previous article on Electric Field.

We have discussed-

• Electric field is the space around an electric charge in which its effect can be experienced.
• Every charge creates an electric field in the space around itself.
• This electric field extends till infinity.

There are following 4 types of electric field-

## Electric Field Lines-

An electric field line or Electric Line of Force may be defined as-

 It is defined as imaginary line in an electric field, the tangent to which at any point on it represents the direction of electric field at that point. OR It is defined as the imaginary path traversed by a unit positive charge in an electric field. OR It is defined as the curve along which a unit positive charge tends to move if free to do so in an electric field.

It is important to note that-

• The electric lines of force are imaginary curves.
• But the electric field represented by them is real.
• The term “lines of force” is misnomer.
• It is more appropriate to call these lines as “electric field lines”.

## Properties of Electric Field Lines-

Following are some of the important properties of electric lines of force-

### Property-01:

• The lines of force are continuous smooth curves.
• They do not have any breaks.

### Property-02:

• The lines of force originate out of a positive charge and terminates on a negative charge.
• If there exists a single charge, then lines of force end at infinity.
• The lines of force cannot form closed loops.

### Property-03:

• The tangent to a line of force at any point gives the direction of electric field at that point.

### Property-04:

• No two lines of force can cross each other.

### Property-05:

• The lines of force have a tendency to contract longitudinally.
• The lines of force have a tendency to exert a lateral pressure on neighboring lines of force.

### Property-06:

• The lines of force are always normal to the surface of a conductor on which the charges are in equilibrium.

### Property-07:

• The relative closeness of the lines of force gives a measure of the strength of the electric field.
• Closer the lines of force, stronger is the electric field.

### Property-08:

• The lines of force do not pass through a conductor.

## Electric Lines Of Force For Different Charged Conductors-

Let us represent the electric lines of force for the following-

• Electric lines of force of a positive point charge
• Electric lines of force of a negative point charge
• Electric lines of force of an electric dipole
• Electric lines of force for a system of two equal positive point charges

### Case-01: For Positive Point Charge-

If there exists only one positive point charge, then-

• The electric lines of force originate from that charge.
• The electric lines of force terminate at infinity.

### Case-02: For Negative Point Charge-

If there exists only one negative point charge, then-

• The electric lines of force originate from infinity.
• The electric lines of force terminate at that charge.

### Case-03: For Electric Dipole-

Electric dipole is a system of two equal and opposite charges separated by a small distance.

If there exists an electric dipole, then-

• The electric lines of force originate from the positive charge.
• The electric lines of force terminate at the negative charge.

### Case-04: For A System Of Two Equal Positive Point Charges-

If there exists a system of two positive charges, then-

• They exert a lateral pressure  on account of repulsion between like charges.
• The electric field is zero at the middle point on the line joining the two charges.
• This point is called neutral point from which no line of force passes.

## Question-01:

Why electric lines of force are smooth continuous curves?

Electric lines of force are smooth continuous curves without any breaks because-

• Their discontinuity would indicate that the charge suddenly vanishes and reappears.
• This is absurd.

## Question-02:

Why no two electric lines of force can intersect each other?

No two electric lines of force can intersect each other because-

• If they intersect, then there will be two tangents at the point of intersection.
• Hence, two directions of the electric field at the same point will exist which is impossible.

## Question-03:

Why electric lines of force contract longitudinally and exerts a lateral pressure?

• Electric lines of force contract length-wise on account of attraction between unlike charges.
• Electric lines of force exert a lateral pressure on account of repulsion between like charges.

## Question-04:

Why electric lines of force are always normal to the surface of a conductor on which the charges are in equilibrium?

This is so because if the lines of force are not normal to the surface of a conductor, then-

• There exists a component of the field $\overrightarrow{E}$ parallel to the surface.
• This would cause the electrons to move and set up a current on the surface.
• But no current flows in the equilibrium condition.

## Question-05:

Why electric lines of force never pass through a conductor?

Electric lines of force never pass through a conductor because electric field inside a charged conductor is zero.

To gain better understanding about Electric Lines of Force,

Watch this Video Lecture

Next Article- Electric Flux

Get more notes and other study material of Class 11 Physics.

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## Electric Field-

Before you go through this article, make sure that you have gone through the previous article on Electric Field.

We have discussed-

• Electric field is the space around an electric charge in which its effect can be experienced.
• Electric field intensity at any point is the strength of electric field at that point.

There are following 4 types of electric fields-

## Important Formulas-

It is important to remember the following formulas while solving problems based on electric field-

### Formula-01:

Electric field intensity (E) at any point is the force (F) experienced by a unit positive charge placed at that point.

The direction of electric field is same as the direction of force.

### Formula-02:

Force experienced by charge q when placed in an electric field having electric field intensity E is given by-

The direction of force is same as the direction of electric field.

### Formula-03:

Acceleration of an electric charge q0 having mass m in an electric field having electric field intensity $\overrightarrow{E}$ is given by-

### Formula-04:

Electric field intensity due to a point charge Q at distance ‘r’ units from it is given by-

### Formula-05:

Electric field intensity at any point P due to the system of N charges is given by-

$\overrightarrow{E}$ = $\overrightarrow{{E}_1}$ + $\overrightarrow{{E}_2}$ + $\overrightarrow{{E}_3}$ + ……. + $\overrightarrow{E{_N}}$

In this article, we will discuss practice problems based on Electric Field.

## Problem-01:

A particle having a charge 1 x 10-9 C experiences a downward force of 1.5 x 10-6 N when placed in an electric field. Calculate the electric field intensity.

## Solution-

Given-

• Charge (q) = 1 x 10-9 C
• Force (F) = 1.5 x 10-6 N

We know,

Electric field intensity (E) is given by-

Substituting the given values, we get-

Electric field intensity (E)

= (1.5 x 10-6 N) / (1 x 10-9 C)

= 1.5 x 103 N/C

Thus, electric field intensity is 1.5 x 103 N/C vertically downward.

## Problem-02:

Calculate the electric field strength required to just support a water drop of mass 10-3 kg and having a charge 1.6 x 10-19 C.

## Solution-

Given-

• Mass of water drop (m) = 10-3 kg
• Charge (q) = 1.6 x 10-19 C

Let electric field intensity required to just support the water drop = E.

To just support the water drop, the condition is-

Force on water drop due to electric field = Weight of water drop

q x E = m x g

Substituting the given values, we get-

1.6 x 10-19 C x E = 10-3 kg x 9.8 m/s2

E = (9.8 x 10-3 kg m/s2) / 1.6 x 10-19 C

E = 6.125 x 1016 N/C

Thus, electric field strength required to just support the water drop is 6.125 x 1016 N/C.

## Problem-03:

An electron enters with a velocity of 5 x 106 m/s along the positive direction of an electric field of intensity 103 N/C. Calculate the time taken by the electron to come temporarily to rest. Mass of the electron is 9.11 x 10-31 kg.

## Solution-

Given-

• Initial velocity of the electron (u) = 5 x 106 m/s
• Electric Field Intensity (E) = 103 N/C
• Mass of the electron (m) = 9.11 x 10-31 kg
• Final velocity of the electron (v) = 0

### Step-01: Calculating Acceleration Of Electron-

We know-

Acceleration of an electric charge q0 having mass m in an electric field having electric field intensity $\overrightarrow{E}$ is given by-

Substituting the given values, we get-

Acceleration of the electron (a)

= (1.6 x 10-19 C x 103 N/C) / (9.11 x 10-31 kg)

= 0.175 x 1015 m/s2

This is actually deceleration which causes the electron to slow down and makes it stop momentarily.

### Step-02: Calculating Time Required To Come To Rest-

By Newton’s first equation of motion, we know v = u + at.

Substituting the given values, we get-

0 = 5 x 106 m/s – 0.175 x 1015 m/s2 x t

0.175 x 1015 m/s2 x t = 5 x 106 m/s

t = (5 x 106 m/s) / (0.175 x 1015 m/s2)

t = 28.57 x 10-9 s

t = 2.9 x 10-8 s

Thus, time taken by the electron to come temporarily to rest is 2.9 x 10-8 seconds.

## Problem-04:

An electron moves a distance of 6 cm when accelerated from rest by an electric field of strength 2 x 104 N/C. Calculate the time of travel. The mass and charge of electron are 9 x 10-31 kg and 1.6 x 10-19 C respectively.

## Solution-

Given-

• Distance traveled by the electron (s) = 6 cm = 0.06 m
• Initial velocity of the electron (u) = 0 m/s
• Electric Field Intensity (E) = 2 x 104 N/C
• Mass of the electron (m) = 9 x 10-31 kg
• Charge on the electron (q) = 1.6 x 10-19 C

### Step-01: Calculating Acceleration Of Electron-

We know-

Acceleration of an electric charge q0 having mass m in an electric field having electric field strength $\overrightarrow{E}$ is given by-

Substituting the given values, we get-

Acceleration of the electron (a)

= (1.6 x 10-19 C x 2 x 104 N/C) / (9 x 10-31 kg)

= 0.36 x 1016 m/s2

### Step-02: Calculating Time Of Travel-

By Newton’s second equation of motion, we know s = ut + ( $\frac{1}{2}$ )at2.

Substituting the given values, we get-

0.06 m = 0 x t + ( $\frac{1}{2}$ ) x 0.36 x 1016 m/s2 x t2

0.12 m = 0.36 x 1016 m/s2 x t2

0.12 s2 = 0.36 x 1016 x t2

t2 = 0.12 / (0.36 x 1016) s2

t2 = 0.33 x 10-16 s2

t = 0.57 x 10-8 s

t = 5.7 x 10-9 s

t = 5.7 ns

Thus, required time of travel is 5.7 nanoseconds.

## Problem-05:

Two point charges of 5 x 10-19 C and 20 x 10-19 C are separated by a distance of 2 m. Find the point on the line joining them at which electric field intensity is zero.

## Solution-

Given-

• Charge (q1) = 5 x 10-19 C
• Charge (q2) = 20 x 10-19 C
• Distance between the two charges = 2 m.

• Electric field due to charge q1 is towards right along the line joining the two charges.
• Electric field due to charge q2 is towards left along the line joining the two charges.

Let-

• Electric field intensity is zero at point P on the line joining the two charges.
• Electric field intensity at point P due to charge q1 is E1.
• Electric field intensity at point P due to charge q2 is E2.
• Distance of point P from charge q1 is x m.
• Distance of point P from charge q2 is (2-x) m.

For electric field to be zero at point P,

Electric field intensity at point P due to q1 = Electric field intensity at point P due to q2

Substituting the given values, we get-

E1 = E2

K x ( $\frac{q_{1}}{x^{2}}$ ) = K x { $\frac{q2}{(2-x))^{2}}$ }

q1 x (2 – x)2 = q2 x x2

5 x 10-19 C x (2 – x)2 = 20 x 10-19 C x x2

(2 – x)2 = 4x2

4 + x2 – 4x = 4x2

3x2 + 4x – 4 = 0

On solving, we get x = 0.67 m.

Thus, electric field is zero at 0.67 meters to the right of charge q1.

## Problem-06:

Two point charges q1 and q2 of 10-8 C and -10-8 C respectively are placed  0.1 m apart. Calculate the electric fields at points A, B and C as shown in figure-

## Solution-

### Calculating Electric Field Intensity At Point A-

Let-

• Electric field intensity at point A due to charge q1 = E1A
• Electric field intensity at point A due to charge q2 = E2A

It is clear that-

• Electric field at point A due to charge q1 is directed towards right.
• Electric field at point A due to charge q2 is directed towards right.

Thus, Net electric field intensity at point A

= |E1A| + |E2A|

= K x ( $\frac{q_{1}}{0.05^{2}}$ ) + K x ( $\frac{q_{2}}{0.05^{2}}$ )

= ( $\frac{K}{0.0025}$ ) x (q1 + q2)

= (K x 400) x (10-8 C + 10-8 C)

= K x 400 x (2 x 10-8 C)

= K x 800 x 10-8 C

= 9 x 109 x 800 x 10-8

= 72000

= 7.2 x 104 N/C

Thus, electric field intensity at point A is 7.2 x 104 N/C directed towards right.

### Calculating Electric Field Intensity At Point B-

Let-

• Electric field intensity at point B due to charge q1 = E1B
• Electric field intensity at point B due to charge q2 = E2B

It is clear that-

• Electric field at point B due to charge q1 is directed towards left.
• Electric field at point B due to charge q2 is directed towards right.

Thus, Net electric field intensity at point B

= |E1B| – |E2B|

= K x ( $\frac{q_{1}}{0.05^{2}}$ ) – K x ( $\frac{q_{2}}{0.15^{2}}$ )

= K x { $\frac{10^{-8}}{0.0025}$$\frac{10^{-8}}{0.0225}$ }

= K x 10-8 x (400 – 44.44)

= 9 x 109 x 10-8 x 355.56

= 32000

= 3.2 x 104 N/C

Thus, electric field intensity at point A is 7.2 x 104 N/C directed towards left.

### Calculating Electric Field Intensity At Point C-

Let-

• Electric field intensity at point C due to charge q1 = E1C
• Electric field intensity at point C due to charge q2 = E2C

It is clear that-

• Electric field at point C due to charge q1 is directed away from q1 along the line joining q1 and C.
• Electric field at point C due to charge q2 is directed towards q2 along the line joining q2 and C.
• Angle between $\overrightarrow{E_{1C}}$ and $\overrightarrow{E_{2C}}$ is 120°.

Now,

Electric field intensity at point C due to charge q1

= K x ( $\frac{q_{1}}{0.1^{2}}$ )

= K x 10-8 x 100

= 9 x 109 x 10-8 x 100

= 9000 N/C directed away from q1 along the line joining q1 and point C.

Similarly,

Electric field intensity at point C due to charge q2

= 9000 N/C directed towards q2 along the line joining q2 and point C.

Net electric field at point C

= sqrt { E21C + E22C + 2 x E1C x E2C x cos120° }

= sqrt { (9000)2 + (9000)2 + 2 x 9000 x 9000 x ( $\frac{-1}{2}$ ) }

= sqrt { 2 x 81 x 106 – 81 x 106 }

= sqrt { 81 x 106 }

= 9 x 103 N/C

Thus, electric field intensity at point C is 9 x 103 N/C.

The direction of electric field is along the bisector of the angle between $\overrightarrow{E_{1C}}$ and $\overrightarrow{E_{2C}}$ towards right.

This is because $\overrightarrow{E_{1C}}$ and $\overrightarrow{E_{2C}}$ are equal in magnitude.

To practice more problems based on Electric Field & Electric Field Intensity,

Watch this Video Lecture

Next Article- Electric Field Lines

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## Electric Field-

Electric Field is defined as-

 The space around an electric charge in which its effect can be experienced is called as its electric field. OR It is the space around an electric charge in which any other charge experiences the coulomb force.

• The charge which produces the electric field is called as a source charge.
• The charge which tests the effect of source charge is called as a test charge.

## Types of Electric Field-

There are following 4 types of electric field-

1. Uniform Electric Field
2. Non-Uniform Electric Field
3. Variable Electric Field
4. Constant Electric Field

### 1. Uniform Electric Field-

It is an electric field at every point of which a unit positive charge experiences the same electric force.

OR

It is an electric field at every point of which the intensity of electric field remains constant in magnitude and direction.

In uniform electric field, the electric lines of force are parallel and equidistant.

### 2. Non-Uniform Electric Field-

It is an electric field at different points of which a unit positive charge experiences the different electric force.

OR

It is an electric field at different points of which the intensity of electric field is different in magnitude and direction.

In non-uniform electric field,  the electric lines of force are non-parallel and non-equidistant.

### 3. Variable Electric Field-

The electric field which changes with respect to time is called as variable electric field.

E = f(t)

### 4. Constant Electric Field-

The electric field which does not depend upon time is called as constant or non-variable electric field.

E ≠ f(t)

## Electric Field Intensity-

Electric Field Intensity is defined as-

 Electric field intensity at any point is the strength of electric field at that point. OR Electric field intensity at any point is the force experienced by a unit positive charge placed at that point.

If $\overrightarrow{F}$ is the force acting on a small test charge q0 at any point in the electric field, then electric field intensity $\overrightarrow{E}$ at that point is given by-

It is important to note that-

• The test charge q0 may have its own electric field.
• This may modify the electric field of the source charge.
• So to minimize this effect, we write $\overrightarrow{E}$ as-

From the above relation,

Force experienced by any point charge q0 in the electric field of intensity $\overrightarrow{E}$ may be calculated as-

$\overrightarrow{F}$ = q0 . $\overrightarrow{E}$

We know F = ma.

So, acceleration of electric charge q0 having mass m in electric field $\overrightarrow{E}$ may be given as-

## Electric Field Intensity Due To Point Charge-

Consider-

• A point charge Q is placed at any point O.
• Electric field intensity due to charge Q has to be calculated at any point P.
• The distance between point O and point P is r units.

To calculate the electric field intensity at point P, we place a point charge q0 at point P.

According to Coulomb’s law,

Force $\overrightarrow{F}$ at point P is given by-

Also, we know $\overrightarrow{E}$ = $\overrightarrow{F}$ / q0

So,

This formula gives the electric field intensity due to a point charge Q at distance ‘r’ units from it.

This formula shows-

• Electric field intensity due to any point charge is inversely proportional to the square of the distance from it.
• As the distance from charge increases, its electric field intensity decreases.
• Electric field intensity becomes zero at infinity.

## Electric Field Intensity Due To System Of Point Charges-

According to principle of superposition of electric charges,

Electric field intensity at a point due to a system of charges is the vector sum of the field intensities due to individual point charges.

Hence, the electric field intensity at any point P due to the system of N charges is-

$\overrightarrow{E}$ = $\overrightarrow{{E}_1}$ + $\overrightarrow{{E}_2}$ + $\overrightarrow{{E}_3}$ + ……. + $\overrightarrow{E{_N}}$

## Important Points-

### Point-01:

• Every charge creates an electric field in the space around itself.
• This electric field extends till infinity.
• Electric field due to a point charge at its own location is undefined.

### Point-02:

• Electric field is a vector quantity since it possess both magnitude and direction.
• The electric field due to a positive point charge is radially outwards away from the charge.
• The electric field due to a negative point charge is radially inwards towards the charge.

### Point-03:

• Electric field intensity is a vector quantity since it possess both magnitude and direction.
• The direction of $\overrightarrow{E}$ is same as that of $\overrightarrow{F}$.
• The SI unit of electric field intensity is N/C (Newton per Coulomb).
• The dimensions of electric field intensity are [MLT-3A-1].

### Point-04:

The value of electric field intensity due to point charge at any point P depends on-

• Magnitude of the charge
• Distance of point P from the charge

It does not depend upon the test charge.

### Point-05:

• Positive charges like alpha, proton, deutron etc experiences forces in the direction of the electric field.
• Negative charges like electron experiences force in the direction opposite to the electric field.
• Neutron experiences no force.

### Point-06:

A point charge q0 is kept in an electric field of strength E and experiences a force F, then-

• E < F / q0 if field is diverging
• E = F / q0 if field is uniform
• E > F / q0 if field is converging

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Next Article- Problems On Electric Field

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## Photoelectric Effect-

Before you go through this article, make sure that you have gone through the previous article on Photoelectric Effect.

We have discussed-

• Photoelectric Effect is a phenomenon of emission of electrons from metal surfaces.
• Photoelectric Effect occurs when a light of suitable frequency falls on the metal surface.

## Photoelectric Cell-

 Photoelectric Cell is a device that converts light energy into electrical energy.

Also, it may also be defined as-

• Photoelectric cell is an electron tube or a photo tube.
• Electrons initiating an electric current in this electron tube originate by photoelectric emission.

Let us discuss the construction and working of a photoelectric cell.

## Construction-

• Photoelectric cell consists of an evacuated glass containing anode and cathode.
• The cathode is semi cylindrical.
• The cathode is coated with a photo sensitive alkali metal like rubidium, molybdenum etc.
• This is to reduce the work function.

## Working-

• When light of suitable frequency is incident on the cathode, it emits the photoelectrons.
• These photoelectrons are attracted by the anode which is positively charged.
• This flow of electrons result in a flow of current.
• The flowing current is proportional to the intensity of incident radiation.
• This current may be further increased by filling some inert gas like argon into the glass.
• The photoelectrons emitted by cathode ionize the gas by collision.
• This collision results in increasing the current.
• The positive ions move towards the cathode.
• Negative ions and electrons move towards the anode.

Obviously, there is no conduction in the absence of light in the phototube.

## Applications of Photoelectric Cell-

Photoelectric cell has its applications in the following-

1. Burglar alarm
2. Counting machines
3. Street lights
4. Home security system
5. Fire alarm

### 1. Burglar Alarm-

• A burglar alarm is fitted near the lock of almirah.
• A thief comes and makes the light fall on the lock by match stick, torch etc.
• The light is received by the photoelectric cell.
• The flow of electrons start from cathode to anode.
• This completes the circuit and alarm starts ringing.

### 2. Counting Machines-

• Counting machines may be operated by using a photoelectric cell.
• Whenever a currency note obstructs the incident light, flow of current stops for that duration.
• This advances the machine by count one.

### 3. Street Lights-

• Street lights may be operated by using a photoelectric cell.
• The switch of the relay is attached with the photoelectric cell.
• When the intensity of sunlight decreases below the required value, the switch is activated.
• This switch on the street lights automatically.

### 4. Home Security System-

• In a home security system, the light is produced by a source at one end of the circuit.
• This light falls on a photoelectric cell sensor located at some distance away from it.
• The electrons are emitted by the cathode.
• When somebody walks between the light source and the sensor, the path of the beam is blocked.
• The sensor registers a drop in light levels and sends a signal to the control box.
• The control box is connected to either siren or a telephone autodialer.
• The control box on receiving the signal responds accordingly.

### 5. Fire Alarm-

• A photoelectric cell with a required circuit is installed in sensitive places.
• When the fire breaks out, light falls on the photoelectric cell.
• This causes the photoelectric emission which causes the current flow.
• This makes the bell connected to the circuit start to ring.

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## Photoelectric Effect-

 It is a phenomenon of emission of electrons from metal surfaces exposed to light energy of suitable frequency.

• The electrons emitted by the metal surface are called as photoelectrons.
• The electric current constituted by photoelectrons is called as photoelectric current.

## Concept-

• Consider a metal piece having free electrons.
• Suppose an electron leaves the metal piece at the cost of its own energy.
• This makes the metal piece positively charged.
• The electrostatic force of attraction starts acting between the positive charge and the electron.
• This force tries to stop the electron to escape out the electric field of the positive charge.
• To escape out, electron must posses certain amount of energy.

Now, we define the following terms-

### Threshold Energy-

• It is the minimum amount of energy required to liberate the electrons from the metal surface.
• It is also called as work function.
• Work function is different for different metals.
• Its unit is Joule or eV (electron volt).
• For a metal, low work function is desirable.

### Threshold Frequency-

It is the minimum frequency that the light must possess to liberate the electrons from the metal surface.

### Threshold Wavelength-

It is the maximum wavelength that the light can possess to liberate the electrons from the metal surface.

### Expression for Threshold Wavelength-

Let-

• W0 = Work function
• h = Planck’s constant = 6.6 x 10-34 Joule-sec
• ν0 = Threshold frequency
• λ0 = Threshold wavelength
• c = speed of light = 3 x 108 m/sec

Then, we have-

This is the required expression for threshold wavelength.

## Einstein’s Photoelectric Equation-

Consider-

• A photon of energy hν falls on a metal surface.
• The energy of this photon is absorbed by a free electron in the metal.

According to Einstein, this absorbed energy hν is utilized for two purposes-

### Purpose-01:

• A part of energy is used by the electron to overcome the surface barrier.
• This part of energy helps the electron to come out of the metal surface.
• This part of energy is equal to the work function of the metal.

### Purpose-02:

• The remaining part of energy is used in giving a velocity to the emitted photoelectron.
• This part of energy is equal to the kinetic energy of the emitted photoelectron.

According to the law of conservation of energy,

If ν = ν0 (threshold frequency), then-

• The free electron is just emitted from the metal surface.
• But kinetic energy of the electron remains 0.

So, above equation may be written as-

0 = W + 0

∴ hν0 = W

Using W = hν0 in the above equation, we get-

The above equation is known as Einstein’s Photoelectric Equation.

This equation shows that-

• Kinetic energy is directly proportional to the frequency.
• Kinetic energy is inversely proportional to the wavelength.

## Important Graphs-

By Einstein’s photoelectric equation,

hν = W + K.E.

∴ K.E. =  hν – W

On comparing with y=mx+c, we get-

m = h and c = -W

The following graph shows the variation of kinetic energy of photoelectron with the frequency of incident light-

For all the metals,

• The slope remains the same.
• This is because slope = h which represents plank’s constant and is same for all.

## Laws of Photoelectric Emission-

The following laws govern the phenomenon of photoelectric emission-

### Law-01:

For a given substance,

• There is a minimum frequency of incident light below which no photoelectric emission occurs.
• This minimum frequency is called as threshold frequency.
• Below threshold frequency, no photoelectric emission occurs whatever the intensity of light may be.

### Law-02:

• Number of photoelectrons emitted per second ∝ Intensity of incident light.
• This is true provided frequency of incident light is greater than threshold frequency.

### Law-03:

• Kinetic Energy of photoelectrons ∝ Frequency of incident light.
• This is true provided frequency of incident light is greater than threshold frequency.
• Kinetic energy of photoelectrons is independent of the intensity of incident light.

### Law-04:

• The process of photoelectric emission is instantaneous.
• As soon as the light of suitable frequency is incident on the substance, it emits the photoelectrons.
• There is no significant delay in emitting the photoelectrons.

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Next Article- Photoelectric Cell

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