Question

In a regular polygon of $10$ sides, each comer is at a distance $R$ from the centre. Identical charges are placed at $9$ corners. At the centre, the magnitude of electric field is $E$ and the potential is $V$. The ratio $\frac{V}{E}$ is

• A. $10 R$
• B. $9 / R$
• C. $\frac{9}{10} R$
• D. $9 R$

Answer 1

Answer: (D)

Since electric potential is a scalar quantity. So, the potential at the centre is given as
$V=K \frac{q}{R}+K \frac{q}{R}+\ldots .9$ times
$V=\frac{9 K q}{R} \ldots$ (i)
Since, electric field is a vector quantity. So, the electric field cancel each other for the charges of opposite corner of polygon.
Only $10 q-(10-1) q=10 q-9 q=q$
will contribute the electric field at the centre of polygon. Thus,
$E=K \frac{q}{R^{2}} \ldots (i) $
From Eqs. (i) and (ii), we get
$\therefore \frac{V}{E}=\frac{\frac{9 K q}{R}}{\frac{K q}{R^{2}}}=9 R$