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Q.

Equipotential surfaces are shown in figure. If the electric field strength will be $4 n V m^{- 1}$ , the value of $n$ is

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Detailed Solution

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Using $d V = - \vec{E} . d \vec{r}$

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$\Rightarrow Δ V = - E . Δ r c o s θ$

$\Rightarrow E = \frac{- Δ V}{Δ r \cos θ}$

$\Rightarrow E = \frac{- (20 - 10)}{10 \times 10^{- 2} \cos 120 °}$

$= \frac{- 10}{10 \times 10 (- \sin 30 °)} = \frac{- 10^{2}}{- 1 / 2} = 200 V / m$

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