Question

In a regular polygon of $n$ sides, each corner is at a distance $r$ from the centre. Identical charges are placed at $(n-1)$ corners. At the centre, the intensity is $E$ and the potential is $V$. The ratio $V/E$ has magnitude.

• A. $rn$
• B. $r(n-1)$
• C. $r$
• D. $r(-1)/n$

Answer 1

Answer: (B)

$\text{Step 1: Electric potential}$

The electric potential is a scalar quantity

So, the potential at the center is the sum of potential due to all $(n-1)$ charges

Potential due to one charge $q$ $=k\frac{q}{r}$

So, Potential for $(n-1)$ charges. i.e, $V=k\frac{(n-1)q}{r}$ $....\left(1\right)$

$\text{Step 2: Electric field}$

The electric field is a vector quantity.

So the electric field cancels each other for the charges of the opposite corners of the polygon. (Refer figure)

For example, suppose $n=8$, there are 3 pairs of equal charges placed at corners so, the electric field of these pairs will be zero (equal and opposite)

Therefore only one charge $q$ will contribute to the electric field at the center of the polygon.

Therefore, electric field $E=k\frac{q}{r^2}$ $....\left(2\right)$

$\text{Step 3: Ratio of electric potential and electric field}$

From equation $\left(1\right)$ and $\left(2\right)$

$\frac{V}{E}=k\frac{(n-1)q}{r}\times \frac{r^2}{kq}=r(n-1)$

Hence, option B is correct