Example 1 A uniform electric field of magnitude E0 and directed along positive x-axis exist in a certain region of space. If at x = 0, then potential V is zero, then what is the potential at x = + x0 ?
Solution: For a uniform field ΔV = EΔr. In this case ΔV = V 0, Δr = x0 0, and E = E0.
Thus V 0 = E (x0 0)
or V = Ex0
Note : The negative sign. As one moves along the direction of electric field, the potential falls
Example 2 Electric field intensity is given by the relation E = 100/x2 where x is in eters. Find potential difference between the points x = 10 m and x = 20 m.
Solution : C = - x1∫x2 = - 10∫20 100x-2dx = 100[x-1]2010
= 100[(1/10)-(1/20)] = 5 volt.
Example 3 In the above question if alternate charges are positive and negative then find potential
x = 0.
Solution: 
Example 4 A metallic charged sphere of radius r. V is the potential difference between point A on the surface and point P distant 4r from the centre of the sphere. Then the electric field at a point which is at a distance 4r from the centre of the sphere will be -
Solution : V = VA VP
= (kq/r)-(kq/4r) = (3kq/4r)
Therefore for point P
E = (kq/(4r)2) = (V/12r)
Example 5 A ring of radius R has a charge +q. A charge q0 is freed from the distance Ö3R on its axis, when it reaches to the centre of the ring, what is its kinetic energy ?
Solution: 
At r = Ö3R , V1 = (kq/2R)
At r = 0 , V2 = (kq/R)
KE = (V2-V1) q0
= (kqq0 / 2R)
Example 6 A solid spherical conductor carries a charge Q. It is surrounded by a concentric uncharged spherical shell. The potential difference between the surface of solid sphere and the shell is V. If a charge of 3Q is given to the shell. Then the new potential difference between the above two points will be
Solution : Initial potential difference before charge is given to the shell
(ii) Final potential difference after the charge 3Q is supplied to the shell
Example 7 In the following diagram the variation of potential with distance r is represented. What is the intensity of electric field in V/m at r = 3m ?
Solution : Because at r = 3m potential V = 5 volt = const.
\ E = - (dv/dr)= 0 V/m
Example 8 In the above example the value of E in V/m at r = 6m will be.
Solution: At r2 = 7m, V2 = 2 volt at r1 = 5m, V1 = 4 volt
Example 9 Electric potential for a point (x, y, z) is given by V = 4x2 volt. Electric field at point (1, 0, 2) is
Solution: E = - (dV/dX) = - 8x = - 8 x 1 = - 8 V/m
\ Magnitude of E = 8 V/m and direction is along.
Example 10 Fig. shows lines of constant potential in a region in which an electric field is present. The value of potentials are written. At which the points A, B and C is the magnitude of the electric field greatest?
Solution : In an electric field, electric intensity E and potential V are related as
For a given line, V = constant and the potential difference between any two consecutive lines dV = V1 V2 = 10 V = const.. So E will be maximum where the distance dr between
the lines is minimum, i.e., at B (where the lines are the closest)
Example 11 A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such
that the surface densities are equal. Find the potential at the common centre.
Solution : If q1 and q2 are the charges on spheres of radii r and R respectively, then q1 + q2 = Q
According to given problem 
Now as potential inside a conducting sphere is equal to its surface, so potential at the
common centre
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