Answer:
The work done is zero.
Solution:
As per the question:
Charge, [tex]q = 1\mu C = 1\times 10^{- 6}\ C[/tex]
Distance moved, d = 5 m
Voltage, V = 6V
Now, we know that an equipotential surface is one where the potential is same everywhere on the surface.
Suppose the the voltage at a distance d = 5 m is V'
Thus
V' = 6 V, (since the surface is equipotential)
Work done in moving a charge is given by:
[tex]W = q\Delta V[/tex]
[tex]W = q(V - V')[/tex]
[tex]W = (1\times 10^{- 6})(V - V')[/tex]
[tex]W = (1\times 10^{- 6})(6 - 6) = 0[/tex]
Thus the work done in moving a charge on an equipotential surface comes out to be zero as the potential difference is zero.
