Along the X-axis, three charges q 2 ,-q and q 2 are placed at x=0, x=a and x=2 a respectively. The resultant electric potential at x=a+r (if a< < r) is : ( ₀ is the permittivity of free space)

Step-by-step solution

We have to find the electric potential at point P . V_P= [ 1 4 ₀ q / 2 (r+a) - 1 4 ₀ q r + 1 4 ₀ q / 2 (r-a) ]= q 4 ₀ [ 1 2(r+a) - 1 r + 1 2(r-a) ]= q 4 ₀ [ r(r-a)-2 (r^2-a^2 )+(r+a) r 2 r (r^2-a^2 ) ]= q 4 ₀ [ r^2-a r-2 r^2+2 a^2+r^2+a r 2 r (r^2-a^2 ) ]= q a^2 4 ₀ r (r^2-a^2 ) Since, r>>a , so we have r^2-a^2 r^2 V_P= q a^2 4 ₀ r^3