Question

The initial charges on the three identical metal spheres in Fig. 21-24 are the following: sphere A, $\mathit{Q}$; sphere B, $\mathbf{-}\mathit{Q}\mathbf{/}\mathbf{4}$; and sphere C, $\mathit{Q}\mathbf{/}\mathbf{2}$, where $\mathit{Q}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}\mathbf{}\mathit{C}$. Spheres Aand Bare fixed in place, with a center-to-center separation of $\mathit{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.20}\mathbf{}\mathit{m}$, which is much larger than the spheres. Sphere Cis touched first to sphere Aand then to sphere Band is then removed. What then is the magnitude of the electrostatic force between spheres Aand B?

Short answer

$F=4.68\times {10}^{-19}N$

Step-by-step solution

Given

Sphere A, Q; sphere B, -Q/4; and sphere C, Q/2, where$Q=2.00\times {10}^{-14}C$

d = 1.20 m

Understanding the concept

UseCoulomb’s law to solve the problem.

Calculate the magnitude of the electrostatic force between spheres A and B

When sphere C touches sphere A, they divide up their total charge (Q/2 + Q) equally between them. Thus, sphere A now has charge 3Q/4, and the magnitude of the force of attraction between A and B becomes

$F=k\frac{(3Q/4)(Q/4)}{d^2}=4.68\times {10}^{-19}N$