Question

Figure 21-37 shows four identical conducting spheres that are actually well separated from one another. Sphere W(with an initial charge of zero) is touched to sphere Aand then they are separated. Next, sphere Wis touched to sphere B(with an initial charge of$\mathbf{-}\mathbf{32}\mathbf{e}$) and then they are separated. Finally, sphere Wis touched to sphere C(with an initial charge of$\mathbf{-}\mathbf{48}\mathbf{e}$), and then they are separated. The final charge on sphere Wis$\mathbf{-}\mathbf{18}\mathbf{e}$.What was the initial charge on sphere A?

Short answer

The initial charge on the sphere A is$+16\mathrm{e}$

Step-by-step solution

The given data

1. Initial Charge of sphere W = 0
2. Initial charge of sphere B = -32e
3. Charge of sphere C = +48e
4. Final charge on sphere W = +18e

Understanding the concept of Coulomb’s law  

According to the properties of a conductor, the charge flows easily through its body. Hence upon making contact with another conductor, the total charge of the system gets split in half between them. This verifies the conservation of charge property. Using this concept, the charge of the sphere can be found.

Calculation of the initial charge of sphere A

As a result of the first action, both sphere W and sphere A possess charge,$\frac{1}{2}{\mathrm{q}}_{\mathrm{A}}$, where$q_A$is the initial charge of sphere A.

As a result of the second action, sphere W has now charge, $\frac{1}{2}\left(\frac{{\mathrm{q}}_{\mathrm{A}}}{2}-32\mathrm{e}\right)$

As a result of the final action, sphere W now has charge equal to:

$\frac{1}{2}\left[\frac{1}{2}\left(\frac{{\mathrm{q}}_{\mathrm{A}}}{2}-32\mathrm{e}\right)+48\mathrm{e}\right]\mathrm{..........................}\left(\mathrm{a}\right)$

Now, we have been given that the final charge of sphere is +18e.

Hence, equating value of final charge from equation (a) to the given final charge, we can get the value of the initial charge of sphere A as:

$\begin{array}{c}\frac{1}{2}\left[\frac{1}{2}\left(\frac{{\mathrm{q}}_{\mathrm{A}}}{2}-32\mathrm{e}\right)+48\mathrm{e}\right]=+18\mathrm{e}\\ {\mathrm{q}}_{\mathrm{A}}=+16\mathrm{e}.\end{array}$

Hence, the value of charge of sphere A is$+16e$