The variation of potential with distance $r$ from a fixed point is shown in Figure. The electric field at $r = 5 cm$ , is :

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Solution

By using the relation,

Electric field intensity, $E = - \frac{d V}{d r}$ (slope of $V - r$ curve)
Where,
Potential difference $= V$ and Distance $= r$

From the given graph,

Electric field intensity, $E = - \frac{(0 - 5)}{(6 - 4)} = - \frac{(- 5)}{2} = E = 2.5 V/cm$

Hence, option (a) is the correct answer.

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Similar questions

The variation of potential with distance $R$ from the fixed point is shown in figure.

The electric field at $R = 5 m$ is

V

r

r = 2 cm

(V)

(r)

(in V/m)

r = 1 m, 3 m

4.5 m

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