Name: Introduction to Electric Charge
Uploaded: 2019-07-19
Duration: 9 min 41 s
Description: In this video, we will study electric charge and its properties and we will also learn about different methods of charging.

Question

Write definition of electric field intensity.

Obtain an expression for electric force and electric pressure on the surface of a charged conductor.

Draw necessary diagram.

Accepted Answer

Electric field intensity :- It is the strength of an electric field at any point. It is equal to the electric force per unit charge experienced by a test charge placed at that point. The unit of it i s volt/meter or Newton/ Coulomb.
Every element of a charged conductor experienced a normal mechanical force. This is the result of repulsive force from similar change present on the rest of the surface of the conductor.
Consider a small element ds on the surface of a charge conductor. if σ is the surface density and the charge carried by the element ds be dq.
Consider a point P just outside the surface near the element ds. The electric intensity at a point P is given by $E = \frac{σ}{\in_{0}}$ .................(1)

The direction of the intensity is normally outwards.

The intensity made up of two components $\vec{E_{1}}$ due to the small change dq present on the element ds.

$\vec{E_{2}}$ due to the remaining charge present on the rest of the surface of the conductor.

Hence, at P, $\vec{E_{1}} + \vec{E_{2}} = \frac{σ}{\in_{0}}$ ..................(2)

Now consider a point Q inside the conductor the electric field $E_{1}$ and $E_{2}$ at this point are oppositely directed. However, electric intensity inside a changed conductor is zero

$\vec{E_{1}} - \vec{E_{2}} = 0$

$\vec{E_{1}} = \vec{E_{2}}$

substituting $E_{1}$ in equation(2), we get,

$2 E_{2} = \frac{σ}{\in_{0}} E_{2} = \frac{σ}{2 \in_{0}}$

Here, $E_{2}$ is the electric field due to the charge on the rest of conductor. Hence, repulsive force experienced by elements ds carrying dq to is given by ,

$\vec{f} = \vec{E_{2}} . d q F = \frac{σ}{\in_{0}} . d q f = \frac{σ}{\in_{0}} . σ . d s F = \frac{σ^{2} d s}{\in_{0}}$

The electric pressure on surface ds will be ,
pressure= force/area

$p r e s s u r e = \frac{σ^{2} d s}{\frac{2 \in_{0}}{d s}} P = \frac{σ^{2}}{2 \in_{0}} N / m^{2}$

Embibe Experts Solutions for Chapter: Electric Charges and Fields, Exercise 1: Rajasthan Board-2019

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Electric Charges and Fields, Exercise 1: Rajasthan Board-2019

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