Of the charge Q initially on a tiny sphere

Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles and are fixed with a certain separation. For what value of q/Q will the electrostatic force between the two spheres be maximized?

Solution:

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The first particle will have a charge of q, and the second particle will have a charge of Q-q. (This is because a portion of q is transferred from the particle that had a charge of Q).

Let $r$ represent the distance between the two particles.

Let us now write Coulomb’s law.

F=k\dfrac{(q_1)(q_2)}{r^2}

Substitute the values of charge into our equation.

F=k\dfrac{(q)(Q-q)}{r^2}

To find the value which maximizes the electrostatic force, we must first take the derivative of our equation and set it to zero.

The derivative of our equation is:

$F(q)=k\dfrac{(q)(Q-q)}{r^2}$

$F'=\dfrac{k}{r^2}(Q-2q)$
(Remember $k$ and $r$ are both constants)

Set it equal to zero to find the maximum.

0=Q-2q

We can now find the ratio.

q=\dfrac{Q}{2}

\dfrac{q}{Q}=0.5

Final Answer:

\dfrac{q}{Q}=0.5

Of the charge Q initially on a tiny sphere

Solution:

Final Answer:

This question can be found in Fundamentals of Physics, 10th edition, chapter 21, question 1.

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