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. What

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.

What must be the distance from sphere A to sphere C on the right such that the original small sphere remains at rest?

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Adela 5 years 2021-08-29T22:47:47+00:00 1 Answers 25 views 0

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  1. Dulcie
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    2021-08-29T22:48:54+00:00
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    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.

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