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In a regular polygon of 10 sides, each corner 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
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the magnitude of electric field is $E$ and the potential is $V$. The ratio $\frac{V}{E}$ is
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Verified Answer
The correct answer is:
$9 R$
Since electric potential is a scalar quantity. So, the potential at the centre is given as
$V=K \frac{q}{R}+K \frac{q}{R}+\ldots .9 \text { times }$
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,
From Eqs. (i) and (ii), we get
$\therefore \quad \frac{V}{E}=\frac{\frac{9 K q}{R}}{\frac{K q}{R^2}}=9 R$
$V=K \frac{q}{R}+K \frac{q}{R}+\ldots .9 \text { times }$
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,
From Eqs. (i) and (ii), we get
$\therefore \quad \frac{V}{E}=\frac{\frac{9 K q}{R}}{\frac{K q}{R^2}}=9 R$
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