Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

(i) Derive an expression for electric field due to a point charge at a distance

r

from the charge and express it in vector form.
(ii) Four identical point charges of

4 μ C

each are placed at the corners of a square of each side

0.1 m

. Calculate the electric field at the centre of the square.
(iii) Calculate the electric field intensity at the centre, when one of the corner charges is removed.

Accepted Answer

(i) Electric field intensity at a point due to a charge is the force acting on a unit positive test charge when placed at the point.
Let us determine force on a charge $q_{\circ}$ placed at P at a distance $r$ from $q$ . It is given by

$\vec{F} = \frac{k q q_{\circ}}{r^{2}} \hat{r}$

Electric intensity is given by

$\vec{E} = \frac{\vec{F}}{q_{\circ}}$

$\Rightarrow \vec{E} = \frac{k q q_{\circ}}{r^{2} q_{\circ}} \hat{r}$

$\Rightarrow \vec{E} = \frac{k q}{r^{2}} \hat{r}$

$\Rightarrow \vec{E} = \frac{1}{4 {πε}_{\circ}} \frac{q}{r^{2}} \hat{r}$ $[∵ k = \frac{1}{4 {πε}_{\circ}}]$

This is the expression for electric field due to a point charge at a distance $r$ from the charge.

(ii) The effective electric field is zero as the electric field by any charge is cancelled by the diagonally opposite charge.

(iii) When one of the corner charges is removed, one pair at opposite corners of value $E$ get cancelled. Due to charge on remaining one corner is
$E = 4 π ε_{\circ} \frac{q \times 1}{r^{2}} = 9 \times 10^{9} \frac{q}{{(\frac{0.1 \sqrt{2}}{2})}^{2}}$

$E = (1.8 \times 10^{11} q) N C^{- 1}$

It is directed towards the corner.

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !