A)
B)
C)
D)
Correct Answer: C
Solution :
As charge on shell B is negative so, potential inside B is constant and negative. Also due to positive charge, potential inside A is positive and constant and greater than that of B. Therefore, as r increases potential decreases and as \[E=KE+PE\] is crossed, potential becomes negative. At the surface of shell B, it is more negative. This variation is represented in graph (c). Note: This problem is considered as wrong, because question asked is for field and graphs are given by potential.You need to login to perform this action.
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