Table of Contents:

• Force Experienced by a Charge in an Electric Field
• Motion of Charged Particle in Electric Field
• Properties of Charged Particle in Electric Field
• Solved Example

Electric Field Intensity is a measure of the force per unit charge at a given point in an electric field.

The electric field around an electric charge is the region where its influence can be felt. The electric field intensity at a point is the force experienced by a unit positive charge placed at that point.

Electric Field Intensity is a vector quantity.

It is denoted by ‘E’.

Formula: Electric Field = $\frac{F}{q}$.

The unit of E is NC^-1 or Vm^-1.

The electric field intensity due to a positive charge is always directed away from the charge and the intensity due to a negative charge is always directed towards the charge.

The electric field intensity at a point d units away from a point charge q is given by the expression:

Electric Field Intensity (E) = $\frac{q}{4\pi ϵd^2}N{C}^{-1}$

The electric field intensity at any point due to a number of charges is equal to the vector sum of the individual field intensities produced by each charge.

Force Experienced by a Charge in an Electric Field

The force experienced by a charge in an electric field is given by, $$\vec{F}=Q\vec{E}$$ where $\overrightarrow{E}$ is the electric field intensity.

Special Cases:

If Q is a positive charge, the force $\overrightarrow{F}$ acts in the direction of $\overrightarrow{E}$. Acceleration $a=\frac{F}{m}=\frac{QE}{m}$.

If Q is a negative charge, the force acts in a direction opposite to $\overrightarrow{E}$. Acceleration $a=\frac{F}{m}=\frac{-QE}{m}$.

A charge in an electric field experiences a force, regardless of whether it is at rest or moving. The electric force is not affected by the mass or velocity of the charged particle; it is only dependent on the charge.

Motion of Charged Particle in Electric Field

The force experienced by a charged particle of charge Q when placed in an electric field of strength E is equal to EQ.

The acceleration of the charged particle in the electric field, $a=\frac{EQ}{m}$.

*The velocity of the charged particle after time t, given an initial velocity of 0, is equal to (EQ/m)t

The distance travelled by the charged particle is $$S = \frac{1}{2}at^2 = \frac{1}{2}\frac{E}{m}t^2 \text{ if the initial velocity is zero.}$$

The Path of a Particle in an Electric Field

When a charged particle is projected into a uniform electric field with some velocity perpendicular to the field, the path traced by it is a parabola. Similarly, when the velocity is not perpendicular to the field, the trajectory of the particle is also a parabola.

When a charged particle of mass m and charge Q remains suspended in a vertical electric field, then mg = EQ.

n = mg/Ee, where m is the mass of the charged particle, Q is the charge, E is the electric field, and e is the fundamental charge.

Pendulum Oscillating in a Uniform Electric Field

The time period of oscillation of a simple pendulum when the bob is given a +ve charge and made to oscillate in a vertically upward electric field is.

Time Period (T) = 2π $\sqrt{\frac{l}{g-\frac{EQ}{m}}}$

In the above case, if Bob is given a negative charge, then the time period is given by;

Time Period (T) = $\frac{2\pi \sqrt{l}}{\sqrt{g+\frac{EQ}{m}}}$

Projectile Motion in an Electric Field

A charged particle with charge Q and initial velocity u projected at an angle to the horizontal in an electric field directed vertically upward will experience a force due to the electric field.

1. Time of Flight = $$\frac{2u \sin \theta}{g \pm \sqrt{\frac{E}{m}}}$$

2. Maximum Height = $$\frac{u^2 \sin^2 \theta}{g \pm \frac{EQ}{m}}$$

3. Range = $$\frac{u^2 \sin 2\theta}{g \pm \frac{E_Q}{m}}$$

Properties of Charged Particles in an Electric Field

1. The electric field density inside a charged hollow conducting sphere is zero.

**2.**The angle θ made by the string with the vertical when a sphere of charge Q is suspended in a horizontal electric field E is given by $$\theta = \tan^{-1}\left(\frac{EQ}{mg}\right)$$.

**3.**The tension in the string is $$\sqrt{EQ^2 + mg^2}$$.

4. The tension in the string of a bob carrying a positive charge, suspended by a silk thread in a vertically upward electric field, is given by: T = mg – EQ.

5. If the bob carries a negative charge, the tension in the string is: $$T = mg + EQ.$$

Force Experienced by Electrons and Protons

The force on the proton is an accelerating force, while the force on the electron is a retarding force. If the proton and electron are initially moving in the direction of the electric field, both particles will experience forces of the same magnitude, but in opposite directions.

Acceleration of proton / Retardation of electron = mass of an electron / mass of a proton.

Solved Example

Question: Given a rigid insulated wireframe in the form of a right-angled triangle ABC set in a vertical plane, with two beads of equal masses m and carrying charges q₁ and q₂ connected by a cord of length L and able to slide without friction on the wires, what is the case when the beads are stationary?

• 1. The angle α (∠APQ)
• 2. The tension in the cord
• 3. The normal reactions of the beads.
• 4. If the cords are now cut, what are the values of the charges for which the beads continue to remain stationary?

Solution:

1. The forces acting on each bead are:

• mg (downward)
• Tension
• Electric Force
• Normal Reactions.

The directions of the forces are illustrated in the figure below:

Equating the forces along and perpendicular to AB and AC to achieve equilibrium of the beads

T cos α = F cos α + mg sin 30℃ (1)

F sin α + N₁ = mg cos 30℃ + T sin α (2)

T sin α = F sin α + mg cos 30℃ (3)

$N^2+Fcos\alpha =Tcos\alpha +mgcos60℃$ (4)

From equations (1) and (3);

(F – T) cos α = mg sin 30℃

T – F sin α = mg cos 30℃

Dividing, cot α = tan 30℃ = cot 60℃

Therefore, α = 60℃.

⇒ Learn More: Equipotential Surface

2. Substituting the value of α in (1) and noting that

\(\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{\ell^2} = F\)

\begin{array}{l}T\ cos\ 60^\circ + mg\ sin\ 60^\circ = \frac{1}{4\pi {{\varepsilon }_{0}}} \frac{{{q}_{1}}{{q}_{2}}}{{{\ell }^{2}}}\end{array}

$$\begin{array}{l}T = \frac{1}{4\pi {{\varepsilon }_{0}}} \frac{{{q}_{1}}{{q}_{2}}}{{{\ell }^{2}}} + mg\end{array}$$

3. N₁ = mg cos 30º + (T - F) sin 60º

⇒ N₁ = mg \begin{array}{l} \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} mg = \sqrt{3} mg \end{array}

N₂ = (T - F) cos 60℃ + mg cos 60℃

2 mg × 1/2 + 2 mg × 1/2 = 4 mg.

4. When the cord is cut, and T = 0, then from equation (1)

0 = F cos 60℃ + mg sin 30℃

F + Mg = 0

This result is based on the assumption that q₁ and q₂ have the same sign, thus indicating a force of repulsion.

Since m_g is fixed in direction, its sign cannot be reversed, but the sign of F can be reversed, since if q₁ and q₂ are of opposite sign, F will change its sign from + to -.

If q₁ and q₂ have opposite signs, then

• F + mg = 0 is the condition for equilibrium.

\begin{array}{l}mg = \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{\ell }^{2}}}\end{array}

$$\begin{array}{l}q_1q_2 = 4\pi\varepsilon_0mg\end{array}$$

Also Read

• Electrostatics
• Electric Potential Energy
• Gauss’s Law
• Coulomb’s Law

Frequently Asked Questions on Electric Field Intensity

Question: Because an infinitesimally small test charge will not disturb the electric field intensity at a point, allowing for an accurate measurement.

The test charge used must be infinitesimally small so that it will not produce a field of its own. The actual value of the electric field intensity will be affected due to the field produced by the test charge.

Question: Electric intensity is scalar or vector quantity?

Electric field intensity is a vector quantity.

Question: Defne electric intensity at a point in an electric field.

The electric intensity at a point is the amount of force exerted on a unit positive charge at that point.