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Q.

The variation of electric potential with distance d from a fixed point is as shown in the figure. The electric intensity at d = 5m is

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a

–2.5 Vm^–1

b

–0.4V m^–1

c

0.4 Vm^–1

d

2.5 Vm^–1

answer is A.

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Detailed Solution

$E = - \frac{dV}{dx} = - \frac{(V_{2} - V_{1})}{(x_{2} - x_{1})} = - \frac{(0 - 5)}{(6 - 4)} = 2 .5 {Vm}^{- 1}$

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