Two tiny spheres carrying charges `1.5 muC` and `2.5 muC` are located 30 cm apart. Find the potential
(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 mu C = 1.5xx10^(-6) C`,
`q_(2) = 2.5 mu C = 2.5xx10^(-6) C`
Distance between the two spheres = 30 cm = 0.3m,
Fig.
(a) `V = ? E = ?` For the mid point 0 , `r_(1) = r_(2) = (0.3)/(2) = 0.15m`
`V = (1)/(4pi in_(0) [(q_(1))/(r_(1)) + (q_(2))/(r_(2))] = (9xx10^(9) (1.5+2.5)xx10^(-6))/(0.15)`
`=2.4xx10^(5) vol t`
(ii) Electric field at 0, `E = E_(2) - E_(1) = (1)/(4 pi in_(0) r^(2)) (q_(2) - q_(1))`
`= (9xx10^(9) (2.5xx10^(6) (2.5xx10^(-6) - 1.5xx10^(6)))/((0.15)^(2)))`
Clearly, `vec(E)` is towards `q_(1)`.
(b) Let p be the point in a plane normal to the line passing through the mid point, where `OP = 10cm = 0.1m, Fig`.
Now `PA = PB = sqrt(0.15^(2) + 0.1^(2) m) = 0.18m`
(i) Potential at P , `V = (q_(1))/(4pi in_(0) (PA)) + (q_(2))/(4pi in_(0) (PB)) = (1)/(4pi in_(0) (PA)) (q_(1) + q_(2))`
`= (9xx10^(9) (1.5xx10^(-6) + 2.5xx10^(-6)))/(0.18) = (9xx10^(9)xx4xx10^(-6))/(0.18) = 2xx10^(5) V`
(ii) Electric intensity at P due to `q_(1) ,`
`E_(1) = (q_(1))/(4pi in_(0) (PA)^(2)) = (9xx10^(9)xx1.5xx10^(-6))/((0.18)^(2)) = 0.42xx10^(6) Vm^(-1)`, along AP produced
Electric intensity at P due to `q_(2)` ,
`E_(2) = (q_(2))/(4pi in_(0) (PB)^(2)) = (9xx10^(9)xx2.5xx10^(-6))/((0.18)^(2)) = 0.69xx10^(6) Vm^(-1)`, along BP produced
From Fig.
`cos (theta)/(2) = (0.10)/(0.18) = 5//9 = 0.5526`
`(theta)/(2) = cos^(-1) 0.5536 = 56.25^(@)` `:. theta = 112.5^(@)`
`cos theta = cos 112.5^(@) = -0.38`
Resulatant field intennsity at P is `E = sqrt(E_(1)^(2) + E_(2)^(2) + 2E_(1) E_(2) cos theta) = 6.58xx10^(5) Vm^(-1)`
Let `alpha` be the angle which resulant intensity `vec(E)` makes `vec(E_(1))`.
`:. tan alpha = (E_(2) sin theta)/(E_(1) + E_(2) cos theta) = (0.69xx10^(6)xx0.9239)/(0.42xx10^(6)+0.69xx10^(6) (-0.38)) (:. sin 112.5^(@) = 0.9239)`.
`tan alpha = (0.688)/(0.16) = 4.3 alpha = tan^(-1) (4.3) = 76.9^(@)`