91影视

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

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.

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, $\frac{1}{2}\left(\frac{{\mathrm{q}}_{\mathrm{A}}}{2}鈭�32\mathrm{e}\right)$

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

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

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:

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

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