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 potential and electric field respectively at O.

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

Where,

∈₀ = Permittivity of free space

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

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 distanceOZ = 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,

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

Electric field due to q at Z,

Electric field due to q₂ at Z,

The resultant field intensity at Z,

Where, 2θis the angle, ∠AZ B

From the figure, we obtain

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.