Question

Sphere B of charge +8q is at a distance a to the left of sphere A, and sphere C of charge +2q is to the right of sphere A.

Answer 1

Answer:

The distance from sphere A to sphere C on the right such that the original small sphere remains at rest is a/2

Explanation:

The force of a charge at a point is given as follows;

From an online source, we have;

E₁ + E₂ = 0

The electric field due to the sphere B of charge +8q = E₁

$E_1 = \dfrac{k \cdot (8 \cdot Q)}{a^2}$

The position of the sphere B = A distance 'a' to the left of 'A'

The electric field due to the sphere C of charge +2q = E₂

The position of the sphere C = A distance to the right of 'A'

Therefore, for the electric field strength of sphere 'B' at 'A', we have;

$E_1 = \dfrac{k \cdot (8 \cdot Q)}{a^2} = 4 \times \dfrac{k \cdot (2 \cdot Q)}{a^2}$

Let 'x' be the distance of the +2q charge to the right of 'A', we have;

$E_2 = \dfrac{k \cdot (2 \cdot Q)}{x^2}$

Therefore, for the force of the +2q charge to balance the +8q charge at C, we have;

$\dfrac{k \cdot (2 \cdot Q)}{x^2} = \dfrac{k \cdot (8 \cdot Q)}{a^2} = 4 \times \dfrac{k \cdot (2 \cdot Q)}{a^2}$

$\therefore \ \dfrac{1}{x^2} = \dfrac{4}{a^2}$

$\therefore \ \sqrt{ \dfrac{1}{x^2} }= \sqrt{\dfrac{4}{a^2}}$

$\dfrac{1}{x} = \dfrac{2}{a}$

$x= \dfrac{a}{2}$

Therefore, the distance, 'x', from sphere A to sphere C on the right such that the original small sphere remains at rest is x = a/2.