Ncert Exercises & Solutions

Q 1.13) Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Distance between the spheres, A and B, r = 0.5 m

Initially, the charge on each sphere $q = 6.5 x 10^{- 7} C$ $q = 6.5 x 10^{- 7} C$

When sphere A is touched with an uncharged sphere C, the total charge on the system of A and C is distributed equally between A and C.

Hence, the final charge on A $= \frac{q + 0}{2} = \frac{q}{2}$ $= \frac{q + 0}{2} = \frac{q}{2}$

When sphere C with charge q/2 is brought in contact with sphere B with charge q, total charges on the system of B and C will divide into two equal halves.

Hence, the final charge on sphere B $= \frac{1}{2} (q + \frac{q}{2}) = \frac{3 q}{4}$ $= \frac{1}{2} (q + \frac{q}{2}) = \frac{3 q}{4}$

Now, the force of repulsion between sphere A and B:

$\begin{array}{l} F = \frac{1}{4 {πε}_{0}} \cdot \frac{q_{A} q_{B}}{r^{2}} = \frac{1}{4 {πε}_{0}} \cdot \frac{\frac{q}{2} \times \frac{3 q}{4}}{r^{2}} \\ = \frac{1}{4 {πε}_{0}} \cdot \frac{3 q^{2}}{8 r^{2}} = \frac{9 \times 10^{9} \times 3 \times {(6 .5 \times 10^{- 7})}^{2}}{8 \times (0 .5)^{2}} = 5 .703 \times 10^{- 3} N \end{array}$ $\begin{array}{l} F = \frac{1}{4 {πε}_{0}} \cdot \frac{q_{A} q_{B}}{r^{2}} = \frac{1}{4 {πε}_{0}} \cdot \frac{\frac{q}{2} \times \frac{3 q}{4}}{r^{2}} \\ = \frac{1}{4 {πε}_{0}} \cdot \frac{3 q^{2}}{8 r^{2}} = \frac{9 \times 10^{9} \times 3 \times {(6 .5 \times 10^{- 7})}^{2}}{8 \times (0 .5)^{2}} = 5 .703 \times 10^{- 3} N \end{array}$

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