question_answer

The variation of potential with distance \[R\] from a fixed point is as shown below. The electric field at \[R=5\,m\]is                                                                     [NCERT 1975; MP PMT 2003]

A)            \[2.5\ volt/m\]

B)                                      \[-2.5\ volt/m\]

C)            \[2/5\ volt/m\]                   

D)            \[-2/5\ volt/m\]

Correct Answer: A

Solution :

           Intensity at 5m is same as at any point between B and C because the slope of BC is same throughout (i.e., electric field between B and C is uniform). Therefore electric field at R = 5m is equal to the slope of line BC hence by \[E=\frac{-dV}{dr}\];                    \[E=-\frac{(0-5)}{6-4}=2.5\frac{V}{m}\]                    At R = 1 m, \[E=-\frac{(5-0)}{(2-0)}=-2.5\frac{V}{m}\]                    and at R = 3m potential is constant so E = 0.