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

 

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

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