Name: Surface and Volume Charge Distribution
Uploaded: 2019-07-19
Duration: 4 min 37 s
Description: This video helps students to understand surface and volume charge distribution. Here, we will also define surface and volume charge densities

Question

Five charges, $q$ each are placed at the corners of a regular pentagon of side 'a'.

(i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the corners (say A) is removed?
(iii) What will be the electric field at O if the charge $q$ at A is replaced by $- q$ ?

How would your answer to be affected if pentagon is replaced by n-sided regular polygon with charge $q$ at each of its corners?

Accepted Answer

Now when Pentagon is replaced by the n-sided polygon
(i) Electric field at $‘ O ’$ will be again zero as all charge distribution about $O$ is symmetric as in the case of the electric field at the centre of conducting ring or shell.
(ii) The electric field at $O$ will remain same as in an (ii) if a charge q is removed. The resultant will be equal to the electric field due to charge $- q$ at $A$ .
So, the electric field at $‘ O ’$ at the centre of a regular polygon of n side if a charge from one vertex is removed is equal to $E_{2} = \frac{q}{4 {πε}_{0} r^{2}}$ . The direction from $A$ to $O$ .
(iii) If charge $- q$ is placed at one vertex after removing $+ q$ from there, then resultant electric field at $‘ O ’$ will be due to charge $(- 2 q)$ at $A$ .
So the net electric field at $O$ after removing $q$ from A by placing $- q$ at A. $E = \frac{- q}{4 {πε}_{0} r^{2}} - \frac{q}{4 {πε}_{0} r^{2}}$
$E = \frac{2 q}{4 {πε}_{0} r^{2}}$ from $O$ to $A$ .

Important Questions on Electric Charges and Fields

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