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Question: The variation of potential with distance R from fixed point is shown in figure. The electric field a...

The variation of potential with distance R from fixed point is shown in figure. The electric field at R = 5m is –

A

2.5 V/m

B

–2.5 V/m

C

2/5 V/m

D

– 2/5 V/m

Answer

–2.5 V/m

Explanation

Solution

OEH is an equipotential surface, the uniform E.F. must be perpendicular to it pointing from higher to lower potential as shown.

Hence E = (i^–j^2)\left( \frac{\widehat{i}–\widehat{j}}{\sqrt{2}} \right)

E = (vE–vB)EB\frac{(v_{E}–v_{B})}{EB} = 0–(–2)2\frac{0–(–2)}{\sqrt{2}}= 2\sqrt{2}

\ E→\overset{\rightarrow}{E} = E. E→\overset{\rightarrow}{E} = 2\sqrt { 2 } (i^–j^)2\frac{(\widehat{i}–\widehat{j})}{\sqrt{2}} = i^–j^\widehat{i}–\widehat{j}