Question

In a regular polygon of $n$ sides, each corner is at a distance $r$ from the centre. Identical charges are placed at $\left( {n - 1} \right)$ corners. At the centre, the intensity is $E$ and the potential is $V$. The ratio $\dfrac{V}{E}$ has magnitude
A. $rn$
B. $r\left( {n - 1} \right)$
C. $r$
D. $\dfrac{{r\left( { - 1} \right)}}{n}$

Answer 1

Hint:To solve this problem, we need to use the formulas for intensity of electric field $E$ and potential $V$ in a regular polygon of $n$ sides, each corner is at a distance $r$ from the centre. After that we will take the ratio of both to find the required answer. While solving this problem, we need to take into consideration the fact that the electric field is a vector quantity and potential is a scalar quantity.

Formulas used:
$V = k\dfrac{{\left( {n - 1} \right)q}}{r}$, $V$ is the potential, $n$ is the number of corners in the polygon, $q$ is the electric charge, $r$is the distance of each corner from the centre and $k$ is a constant with a value of \[8.99 \times 109N{m^2}/{C^2}\].
$E = k\dfrac{q}{{{r^2}}}$, where, $E$ is the intensity or electric field, $q$ is the electric charge, $r$ is the distance of each corner from the centre and $k$ is a constant with a value of \[8.99 \times 109N{m^2}/{C^2}\].

Complete step by step answer:
Here, we are asked to find the magnitude of the ratio $\dfrac{V}{E}$.
We know that $V = k\dfrac{{\left( {n - 1} \right)q}}{r}$ and $E = k\dfrac{q}{{{r^2}}}$
$
\dfrac{V}{E} = \dfrac{{k\dfrac{{\left( {n - 1} \right)q}}{r}}}{{k\dfrac{q}{{{r^2}}}}} \\
\therefore\dfrac{V}{E} = r\left( {n - 1} \right) \\
 $
Thus, the magnitude of the ratio $\dfrac{V}{E}$ is $r\left( {n - 1} \right)$.

Hence, option B is the right answer.

Note:Here, we have taken $V = k\dfrac{{\left( {n - 1} \right)q}}{r}$ because potential is the scalar quantity. Therefore, the potential at the centre is the total sum of potential due to $\left( {n - 1} \right)q$ number of charges. And we have taken $E = k\dfrac{q}{{{r^2}}}$ because it is a vector quantity. Therefore, the electric field cancels each other for the charges of opposite corners of the polygon. Only charge $q$ will be responsible for this electric field or intensity.