Question

Determine magnitude of the electric field at the point P shown in the figure(Figure 1) . The two charges are separated by a distance of 2a. Point P is on the perpendicular bisector of the line joining the charges, a distance x from the midpoint between them.

Determine the direction of the electric field. Assume that the positive x and y axes are directed to the right and upward respectively.

Answer 1

Answer 2

Let d be the distance of P from both the charges.
The magnitude of each force on unit charge at P
= kQ/d^2.
The force at P due to Q is away from Q where as due to -Q is toward -Q

Resolving the forces the horizontal components cancel each other where as the vertical components add.
The angle between one of the forces and the horizontal is given 'by tan ? = (a/x)
The sum of the vertical forces
= 2 (kQ/d^2) sin ?.

d^2 = a^2 + x^2 and sin ? = a / d = a/ ? [a^2 + x^2]

Substituting
The electric field at the point P
= 2 {kQ/ [a^2 + x^2]} *{a/ ? [a^2 + x^2]}
= 2KQa / [a^2 + x^2] ^ (3/2)

Answer 3

E1(due to +Q at P)= kQ/r^2, where r=sqrt(x^2+a^2)

= kQ/(x^2+a^2) in a direction making angle 45 degree clockwise with x axis downward

E2(due to -Q at P)= -kQ/r^2

= -kQ/(x^2+a^2) in a direction making angle 45 degree anticlockwise with x axis downward

E(net at P) = sqrt(E1^2+E2^2+2E1*E2*cos90)

= k/r^2sqrt(Q1^2+Q2^2)

= k/(x^2+a^2)*sqrt(Q1^2+Q2^2) in a dirction at 90 degree downward from x axis.