In a regular polygon of n sides, each corner is at a distance r from centre. Identical charges are p

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2. Electric Potential and Capacitance

hard

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.

$r$ $n$

$r$ $(n - 1)$

$(n - 1)/r$

$r(n - 1)/n$

Solution

The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to $(n-1) q$ number of charges.i.e, $V=k \frac{(n-1) q}{r}$

The electric filed is a vector quantity. So the electric field cancel each other for the charges of opposite corner of polygon. Only charge $n q-(n-1) q=q$ will contribute

the electric field at the center of polygon. thus, $E=k \frac{q}{r^{2}}$

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

Standard 12

Physics

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