Two opposite charges each of magnitude 500 C are 10 ~cm apart. Find electric field intensity at a distance of 25 ~cm from the midpoint on axial line of the dipole.

Step-by-step solution

Given, charge of dipole, q=500 10⁻⁶ C Separation between charges, 2 a=10 ~cm =10 10⁻² ~m Distance of location from mid point of axis, r=25 ~cm =25 10⁻² ~m Electric field on axis of dipole, E= 4 k q a r (r^2-a^2 )^2 where, k is Coulomb's constant =9 10^9 C ^2 ~m ⁻² ~N ⁻¹ aligned & 4 9 10^9 500 10⁻⁶ & E= 5 10⁻² 25 10⁻² [ (25 10⁻² )^2- (5 10⁻² )^2 ]^2 & = 2.25 10^5 [0.0625-0.0025]^2 & = 2.25 10^5 (0.06)^2 & = 2.25 10^5 3.6 10⁻³ & =6.25 10^7 & 6 10^7 NC ⁻¹ & aligned The result is close to option (a).