Get Answers to all your Questions

header-bg qa

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 the pentagon is replaced by an n-sided regular polygon with charge q at each of its corners?

Answers (1)

Explanation:-

a) i) For a regular pentagon with equal charges q placed at each corner, the charges are symmetrically arranged. The electric field due to each charge at the center O cancels out due to symmetry. Each charge creates an electric field at OOO, but their components along the pentagon's symmetry axes cancel out.

The net charge at the center will be zero as all the charges are equidistant from the center 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, the net electric field at the center 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 the 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) As n increases, the symmetry of the arrangement becomes more perfect. In the limit of very large n (approaching a circle), the net electric field at the center O remains. If the pentagon is replaced by a polygon and charges Q at each corner, then also they all will cancel out each other, and the net electric field at O will be zero.

 

 

Posted by

infoexpert21

View full answer