Doubt by Aditya
Hint : Remember the last property of equipotential surfaces.
Solution :
We know,
E = -dv/dr
As we can see from the figure that the potential difference between any two consecutive surfaces is 10 V i.e. dv = 10 V
E = -dv/dr
As we can see from the figure that the potential difference between any two consecutive surfaces is 10 V i.e. dv = 10 V
When dv = const.
E ∝ 1/dr
Clearly, the smaller the distance between two surfaces higher will be the electric field.
Clearly, the smaller the distance between two surfaces higher will be the electric field.
From the figure, it is clear that 20V and 10V surfaces has the least distance hence we can say that electric field between these two surfaces will be the highest.
Hence the point B will have the highest electric field strength.