Consider two equipotential surfaces $A$ and $B$ having different potentials. Then

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$A l i n e o f f o r c e f r o m A t o B m u s t b e p e r p e n d i c u l a r t o b o t h .$

$A a n d B c a n i n t e r s e c t .$

$A a n d B c a n n o t i n t e r s e c t .$

$I n t h e r e g i o n b e t w e e n A a n d B, t h e f i e l d i s m a x i m u m w h e r e e q u i p o t e n t i a l s u r f a c e s a r e c l o s e s t t o e a c h o t h e r .$

answer is A, C, D.

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Detailed Solution

As we know that two equipotential surfaces can not intersect each other and lines of force are always perpendicular to the equipotential surface. Ifthe equipotential surfaces intersect ,then at a point on the line of intersection the electric field will have two different directions which is not possible .

When two equipotential lines close together, the slope will be steep. So, close equipotential surfaces indicate a maximum electric field. $(E = - \frac{△ V}{△ x})$

Hence correct answer is (1), (3), (4)

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