Question

Two tiny spheres carrying charges 1.5 μC and 2.5 μC 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 midpoint 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, q₁ = 1.5 μC

Magnitude of charge located at B, q₂ = 2.5 μC

Distance between the two charges, d = 30 cm = 0.3 m

(a) Let V₁ and E₁ are the electric potentials and electric field respectively at O.

V₁ = Potential due to charge at A + Potential due to charge at B

`"V"_1 = "q"_1/(4piin_0("d"/2)) + "q"_2/(4piin_0("d"/2)) = 1/(4piin_0("d"/2))("q"_1 + "q"_2)`

Where,

∈₀ = Permittivity of free space

`1/(4piin_0) = 9 xx 10^9  "N C"^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"`

E₁ = Electric field due to q₂ − Electric field due to q₁

= `"q"_2/(4piin_0("d"/2)^2) - "q"_1/(4piin_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 × 10⁵ V and the electric field at mid-point is 4 × 10⁵ 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.

V₂ and E₂ are the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

`"BZ" = "AZ" = sqrt((0.1)^2 + (0.15)^2)` = 0.18 m

V₂ = Electric potential due to A + Electric Potential due to B

= `"q"_1/(4piin_0("AZ")) + "q"_2/(4piin_0("BZ"))`

= `(9 xx 10^9 xx 10^-6)/0.18(1.5 + 2.5)`

Electric field due to q at Z,

`"E"_"A" = "q"_1/(4piin_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 q₂ at Z,

`E_"B" = "q"_2/(4piin_0("BZ")^2)`

= `(9 xx 10^9 xx 2.5 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 + 2"E"_"A""E"_"B"cos 2 theta)`

Where, 2θ is the angle, ∠AZB

From the figure, we obtain

`cos theta = 0.10/0.18 = 5/9 = 0.5556`

`theta = cos^-1 0.5556 = 56.25`

∴ `2theta` = 112.5°

`cos2 theta = -0.38`

`"E" = 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 × 10⁵ V and electric field is 6.6 ×10⁵ V m⁻¹.