Question

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.

Answer 1

Two charges placed at points A and B are represented in the given figure. O is the mid point of the line joining the two charges.

Magnitude of charge located at a, `q_(1) = 1.5 mu C`
Magnitude of charge located at B, `q_(2) = 2.5 mu C`
Distance between the two charges, `d = 30 cm = 0.3 m`
(a) Let `V_(1)` and `E_(1)` are the potential and electric field respectively at O.
`V_(1)` = Potential due to charge at A+ Potential due to charge at B
`V_(1) = (q_(1))/(4pi in_(0) ((d)/(2)))+(q_(1))/(4pi in_(0)((d)/(2))) = (1)/(4pi in_(0)((d)/(2)))(q_(1)+q_(2))`
Where,
`in_(0)` = Permittivity of fre space
`(1)/(4pi in_(0)) = 9 xx 10^(9) NC^(2)m^(-2)`
`:. V_(1) = (9 xx 10^(9) xx 10^(-6))/(((0.30)/(2))) (2.5 + 1.5) = 2.4 xx 10^(5)V`
`E_(1)` = Electric field due to `q_(2)` - Electric field due to `q_(1)`
`= (q_(1))/(4pi epsilon_(0)((d)/(2))^(2))-(q_(1))/(4pi epsilon_(0)((d)/(2))^(2))`
`= (9 xx 10^(9))/((0.30)/(2))^(2) xx 10^(6) xx (2.5 - 1.5)`
`= 4 xx 10^(5) V m^(-1)`
Therefore, the potential at mid-point is `2.4 xx 10^(5)V` and the electric field at mid-point is `4 xx 10^(5) V m^(-1)`. The field is directed from the larger charge to the smaller charge.
(b) consider a point Z such that normal distance OZ = 10 cm = 0.1 m, as shown in the following figure.

`V_(2)` and `E_(2)` are the electric potential and electric field respectively at Z.
It can be observed from the figure distance,
`BZ = AZ = sqrt((0.1)^(2)+(0.15)^(2)) = 0.18 m`
`V_(2)` = Electric potential due to A + Electric Potential due to B
`= (q_(1))/(4pi epsilon_(0)(AZ))+(q_(1))/(4pi epsilon_(0)(BZ))`
`= (9 xx 10^(9) xx 10^(-6))/(0.18) (1.5+2.5)`
`= 2 xx 10^(5)V`
Electric field due to q at Z,
`E_(A) = (q_(1))/(4pi epsilon_(0) (AZ)^(2))`
`= (9 xx 10^(9) xx 1.5 xx 10^(-6))/((0.18)^(2))`
`= 0.416 xx 10^(6) V//m`
Electric field due to `q_(2)` at Z,
`E_(B) = (q_(2))/(4pi epsilon_(0)(BZ)^(2))`
`= (9 xx 10^(9) xx 10^(-6))/((0.18)^(2))`
`= 0.69 xx 10^(6) V m^(-1)`
The resultant field intensity at Z,
`E = sqrt(E_(A)^(2)+E_(B)^(2)+2E_(A)E_(B) cos 2 theta)`
Where, `2 theta` the angle `/_ AZ B`
From the figure, we obtain
`cos theta = (0.10)/(0.18) = (5)/(9) = 0.5556`
`theta = cos^(-1@)0.5556 = 56.25`
`:. 2 theta =112.5^(@)`
`cos 2theta = -0.38`
`sqrt((0.416 xx 10^(6))^(2) xx (0.69 xx 10^(6))^(2)+ 2 xx 0.416 xx 0.69 xx 10^(12) xx(0.38))`
`= 6.6 xx 10^(5) V m^(-1)`
Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is `2.0 xx 10^(5) V` and electric field is `6.6 xx 10^(5)V m^(-1)`.