Five charges, q each are placed at the corners of a regular pentagon of side 'a' (Fig. 1.12).

 

(a)
(i) What will be the electric field at O, the center 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?


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

Answers (1)

i) The net charge at the centre will be zero as all the charges are equidistant from the centre and they will cancel out each other.

ii) It can be written as that the electric field at O due to A and all other charges is zero. The equation would be as follows:
\vec{E}_A+\vec{E}_{f\! our\: charges}=0
Hence \vec{E}_{f\! our\: charges}=-\vec{E}_A\: \: or \left | \vec{E}_{f\! our\: charges} \right |= \left |\vec{E}_A \right |

When charge q is removed from A, net electric field at the centre due to
remaining charges \left | \vec{E}_{f\! our\: charges} \right |= \left |\vec{E}_A \right |=\frac{1}{4\pi \varepsilon _0}\frac{q}{r^{2}}\: \: along\: \: OA
 

(iii) If charge q at A is replaced by -q, then electric field due to this negative charge

\vec{E_{-q}}=\frac{1}{4\pi \varepsilon _0}\frac{q}{r^{2}}\: along \: OA
Hence net electric field at the centre
\vec{E_{net}}=\vec{E_{-q}}+\vec{E}_{force \: charge}=\frac{1}{4\pi \varepsilon _0}\frac{q}{r^{2}}\: +\frac{1}{4\pi \varepsilon _0}\frac{q}{r^{2}}\:

\vec{E}_{net}=\frac{1}{4\pi \varepsilon _0}\frac{2q}{r^{2}}\: \: along \: OA

b) If pentagon is replaced by a polygon and charge Q at each corner, then also they all will cancel out each other, and the net electric field at O will be zero.

 

 

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