Question

In Fig. 21-24, three identical conducting spheres initially have the following charges: sphere A,$\mathbf{4}\mathbf{Q}$ ; sphere B,$\mathbf{-}\mathbf{6}\mathbf{Q}$ ; and sphere C,$\mathbf{0}$ . Spheres Aand Bare fixed in place, with a center-to-center separation that is much larger than the spheres. Two experiments are conducted. In experiment 1, sphere Cis touched to sphere Aand then (separately) to sphere B, and then it is removed. In experiment 2, starting with the same initial states, the procedure is reversed: Sphere Cis touched to sphere Band then (separately) to sphere A, and then it is removed. What is the ratio of the electrostatic force between Aand Bat the end of experiment 2 to that at the end of experiment 1?

Short answer

The ratio of the electrostatic force between A and B at the end of experiment 2 to that of at the end of experiment 1 is$\mathrm{0.375.}$

Step-by-step solution

The given data

The charge of sphere A is$4Q$

The charge of sphere B is$-6Q$

The charge of sphere C is$0$

The separation between them is larger than spheres.

Experiment 1: Sphere C is touched to sphere A and then to sphere B and then removed

Experiment 2: Sphere C is touched to sphere B and then to sphere a and then removed.

Understanding the concept of Coulomb’s law

According to Coulomb's Law of electrostatic attraction or repulsion within particles, the force acting on them is given as being directly proportional to the product of the charges on the particles and being inversely proportional to the separation between them. Using this concept, we can find out the force acting on them.

Formula:

The magnitude of the electrostatic force between any two particles,

${\mathrm{F}}_1=\mathrm{k}\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$ (1)

Calculation of the ratio of force between A and B for experiment 2 to experiment 1 

In experiment 1, sphere C first touches sphere A, and they divided up their total charge (Q/2 plus Q) equally between them. Thus, sphere A and sphere C each acquired charge 3Q/4. Then, sphere C touches B and those spheres split up their total charge (3Q/4 plus –Q/4) so that B ends up with a charge equal to Q/4. The force of repulsion between A and B using equation (1) is given as:

${\mathrm{F}}_1=\mathrm{k}\frac{\left(\frac{3\mathrm{Q}}{4}\right)\left(\frac{\mathrm{Q}}{4}\right)}{{\mathrm{d}}^2}\mathrm{.............................}\left(2\right)$

Now, in experiment 2, sphere C first touches B, which leaves each of them with charge Q/8. When C next touches A, sphere A is left with charge 9Q/16. Consequently, the force of repulsion between A and B using equation (1) is given as:

${\mathrm{F}}_2=\mathrm{k}\frac{\left(\frac{9\mathrm{Q}}{16}\right)\left(\frac{\mathrm{Q}}{8}\right)}{{\mathrm{d}}^2}\mathrm{..............................}\left(3\right)$

Thus, the required ratio of the force between A and B for experiment 2 to experiment 1 is given by dividing equations (3) by (2) as follows:

$\begin{array}{c}\frac{{\mathrm{F}}_2}{{\mathrm{F}}_1}=\frac{\left(\frac{9}{16}\right)\left(\frac{1}{8}\right)}{\left(\frac{3}{4}\right)\left(\frac{1}{4}\right)}\\ =0.375\end{array}$

Hence, the value of the ratio is $\mathrm{0.375.}$