Respuesta :
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
[tex]E_1 = \dfrac{k \cdot (8 \cdot Q)}{a^2}[/tex]
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;
[tex]E_1 = \dfrac{k \cdot (8 \cdot Q)}{a^2} = 4 \times \dfrac{k \cdot (2 \cdot Q)}{a^2}[/tex]
Let 'x' be the distance of the +2q charge to the right of 'A', we have;
[tex]E_2 = \dfrac{k \cdot (2 \cdot Q)}{x^2}[/tex]
Therefore, for the force of the +2q charge to balance the +8q charge at C, we have;
[tex]\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}[/tex]
[tex]\therefore \ \dfrac{1}{x^2} = \dfrac{4}{a^2}[/tex]
[tex]\therefore \ \sqrt{ \dfrac{1}{x^2} }= \sqrt{\dfrac{4}{a^2}}[/tex]
[tex]\dfrac{1}{x} = \dfrac{2}{a}[/tex]
[tex]x= \dfrac{a}{2}[/tex]
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.
