Two tiny spheres carrying charges `1.5 muC and 2.5 muC` are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this mid-point in a plane normal to the line and passing through the mid point.

Explanation

Here `q_1 = 1.5 muC = 1.5 xx 10^(-6) C, q_2 = 2.5 muC = 2.5 xx 10^(-6)C`, distance `AB = 30 cm = 0.3 m`
(a) At mid -point O, `AO = OB = 0.3/2m= 0.15 m`
As potential is a scalar and both charges are +ve, hence resultant potential at point O
`V_0 = V_1 + V_2 = 1/(4pi epsi_0). [q_1/(AO) + q_2/(OB)]`
`=9 xx 10^9 [(1.5 xx 10^(-6))/(0.15) + (2.5 xx 10^(-6))/(0.15)`
`= 2.4 xx 10^5 V`.
Electric fields at point O due to two charges are in mutually opposite directions, hence resultant field is the difference of fields due to two charges.

`E_0 = E_2 - E_1` along `BOA = 1/(4pi epsi_0). [q_2/((OB)^2)- q_1/((OA)^2)]`
`=9 xx 10^9 [(2.5 xx 10^(-6))/((0.15)^2) - (1.5 xx 10^(-6))/((0.15)^2)] = 4.0 xx 10^5 N C^(-1)` along BOA.
At a point P situated at the line normal to the axis and passing through mid-point O, where
OP = 10 cm = 0.1m, we have `AP = BP sqrt((0.15)^2 + (0.1)^2) = sqrt(0.0325)m`
`:.` Total potential at point P
`V = 1/(4 pi epsi_0)[q_1/(AP) + q_2/(BP)] = 9 xx 10^(9) [ (1.5 xx 10^(-6))/sqrt(0.0325) + (2.5 xx 10^(-6)/sqrt(0.0325)] = 2.0 xx 10^5 V`.
Again electric field at P due to `q_1`
`E_2 = 1/(4pi epsi_0) . q_1/((AP)^2) = (9 xx 10^9 xx 1.5 xx 10^(-6))/((0.0325)) = 4.15 xx 10^5 N C^(-1) ` along AP and electric field at P due to `q_2`
`E_2 = 1/(4pi epsi_0). q_2/((BP)^2) = (9 xx 10^9 xx 2.5 xx 10^(-6))/((0.0325)) = 6.92 xx 10^5 N C^(-1)` along BP
Let angle `angle APB = theta," then " angle APO = angle OPB = theta/2` such that
`tan"" (theta)/2 = (OB)/(OP) = 0.15/0.1 = 1.5 :. theta/2 = tan^(-1) (1.5) = 56.3^@ or theta = 2 xx (56.3^@) = 112.6^@`
`:.` Total electric field at point P will be `E = sqrt(E_1^2 + E_2^2 + 2E_1 E_2 cos theta)`
`=sqrt((4.15xx10^(5))^2 +(6.92 xx 10^5)^2 + 2 xx (4.15 xx 10^5) xx (6.92 xx 10^5)xx cos 112.6^@)`
`=6.56 xx 10^5 N C^(-1) = 6.6 xx 10^(5) NC^(-1)`
This field E is making an angle a from direction of `E_1`, where
`tan alpha = (E_2 sin theta)/(E_1 + E_2 costheta) = (6.92 xx 10^5 sin 112.6^@)/(4.15 xx 10^5 + 6.92 xx 10^5 cos 112.6^@) = 4.2858`
`:. alpha = tan^(-1) (4.2858) = 76.9^@`
`:.` Angle subtended by `vec E ` from the line AB
`beta = 90^@ - (alpha - theta/2) = 90^@ - alpha + theta/2 = 90^@ - 76.9^@ + (112.6^@)/2 = 69.4^@ = 69^@`