Solved Problem on Electric Charge

A sphere A, initially charged with 8 coulombs, is brought into contact with another sphere B, charged with −2 coulombs. After being separated, sphere B is grounded. Next, another sphere C, initially charged with −5 coulombs, is brought into contact with sphere A, and then they are separated. Sphere A is then brought into contact with sphere B again and then separated. Determine the final charges on each sphere. Assume that all three spheres are made of the same material and have the same size.

Problem data:

Charge of sphere A: Q_A = 8 C;
Charge of sphere B: Q_B = −2 C;
Charge of sphere C: Q_C = −5 C.

Solution:

Initially, we have the following situation

Figure 1

When spheres A and B are brought into contact, the total charge will be equally distributed between them, and the final charge of each sphere will be the average of the two initial charges.

\[ \begin{gather} Q_{\small A}=Q_{\small B}=\frac{8\;\mathrm C+(-2\;\mathrm C)}{2}=\frac{8\;\mathrm C-2\;\mathrm C}{2}=\frac{6\;\mathrm C}{2}=3\;\mathrm C \end{gather} \]

Figure 2

When sphere B is grounded, it discharges completely, and its charge is equal to zero.

\[ \begin{gather} Q_{\small B}=0 \end{gather} \]

Figure 3

When spheres A and C are brought into contact, the total charge will again be equally distributed between them. The final charge of each sphere will be the average between the spheres, and the charge of the sphere is the charge acquired after contact with B ( \( Q_{\small A}=3\;\mathrm C \) )

\[ \begin{gather} Q_{\small A}=Q_{\small C}=\frac{3\;\mathrm C+(-5\;\mathrm C)}{2}=\frac{3\;\mathrm C-5\;\mathrm C}{2}=-\frac{2\;\mathrm C}{2}=-1\;\mathrm C \end{gather} \]

Figure 4

When spheres A and B are brought into contact again, the total charge will be equally distributed between them. The charge of sphere A used here is the one after its contact with C ( \( Q_{\small A}=-1\;\mathrm C \) ), and the charge of sphere B is zero after being grounded ( \( Q_{\small B}=0 \) )

\[ \begin{gather} Q_{\small A}=Q_{\small B}=\frac{-1\;\mathrm C+0}{2}=-\frac{1\;\mathrm C}{2}=-0.5\;\mathrm C \end{gather} \]

Figure 5

The final charges will be:

Charge of the sphere A: Q_A = − 0.5 C;
Charge of the sphere B: Q_B = − 0.5 C;
Charge of the sphere C: Q_C = − 1 C.

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