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 magnitude of intensity is E and the potential is V. The ratio V/E is

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 magnitude of intensity is E and the potential is V. The ratio V/E is (A) rn (B) r (n-1) (C) (n-1)/r (D) r(n-1) /n

Tardigrade Admin · Accepted Answer

At the centre, the intensity is effectively due to one charge and the potential is due to (n - 1) charges ∴ E=(KQ/r2) and V=(k(n-1)Q/r) Hence, (V/E)=(k(n-1)Q/r)×(r2/KQ) =(n-1)r

Q. 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 magnitude of intensity is $E$ and the potential is $V$ . The ratio $V / E$ is

$r n$

$r (n - 1)$

$(n - 1) / r$

$r (n - 1) / n$

At the centre, the intensity is effectively due to one charge and the potential is due to $(n - 1)$ charges
$∴ E = \frac{K Q}{r ^{2}}$ and $V = \frac{k ( n - 1 ) Q}{r}$
Hence, $\frac{V}{E} = \frac{k ( n - 1 ) Q}{r} \times \frac{r ^{2}}{K Q}$
$= (n - 1) r$

Q. 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 magnitude of intensity is E and the potential is V. The ratio V/E is

rn

r(n−1)

(n−1)/r

r(n−1)/n