Question

(a) Derive and expression for the electric field at any point on the equatorial line of an electric dipole.
(b) Two identical point charges, q each kept 2m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.

Answer 1

(a) Electric field intensity at point P due to + ve q charge at point P

`|vec(E)_(+q)|=1/(4piin_(0))q/((r^(2)+a^2))" ...(1)"`
Electric field intensity at point P due to - ve q charge at point P
`|vec(E)_(-q)|=1/(4piin_(0))q/((r^(2)+a^2))" ...(2)"`
`"Hence, "|vec(E)_(+q)|=|vec(E)_(-q)|`
`E_(+q) sin theta and E_(-q) sin theta` are equal in magnitude and oppsite in direction so they cancel out.
Net electric field at point P
`E=2E_(+q)costheta`
`E=2""1/(4piin_(0))q/((r^(2)+a^(2)))a/((r^(2)+a^(2))^(1//2))`
`E=1/(4piin_(0))(2(2a))/((r^(2)+a^(2))^(3//2))`
`P=q(20)`
Where P is electric dipole monent
`E=1/(4piin_(0))1/((r^(2)+a^(2))^(3//2))`
`"Vetor form "vec(E)=1/(4piin_(0))1/((r^(2)+a^(2))^(3//2))`
(b) As charge `theta` is in equilibrium net force on the charge `theta` is zero
`1/(4piin_(0))-1/(4piin_(0))(qQ)/((2m-x)^(2))=0`

`1/(4piin_(0))(qQ)/x^(2)=1/(4pi)(qQ)/((2m-x)^(2))`
`x=2m-x`
`2x=2m`
`2=m`
Hence, Q is at the mid point of the line joining the charge q and q.
as the whole system is also in equilibrium, force an the charge q must also be zero.
`1/(4piin_(0))(qQ)/x^(2)+1/(4piin_(0))(qq)/((2m)^(2))=0`
`1/(4piin_(0))(qQ)/x^(2)+1/(4piin_(0))(qq)/((2m)^(2))=0`
`Q/(x^(2))=(-q)/(4m^(2))" as "x=m`
`Q=(-q)/4`