Question

Two equal and opposite charges 6.0 μC and -6.0 μC are separated by a distance of 2 cm. The electric potential at a point 30 cm away from the axis of the dipole is ?

Answer 1

Since Electric potential due to a dipole at its axis is given by,

V = k*p/(r^2 - d^2)

here, k = 9*10^9

p = dipole moment = q*d

q = 6.0 C = 6.0*10*-6 C

r = axial distance = 30 cm = 0.3 m

d = distance between charges = 2 cm = 0.02 m

Since, r>>>d

So, r^2 - d^2 r^2

then, V = k*p/r^2 = (9*10^9)*(6.0*10^-6)*(0.02)/(0.3^2)

V = 12000 volt

Please upvote.