Ncert Exercises & Solutions

Q 1.12) (i) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{- 7} C$ ? The radii of A and B are negligible compared to the distance of separation.

(ii) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

(a)

Charge on the sphere $A, q_{A} = 6 .5 \times 10^{- 7} C$
Charge on sphere $B, q_{B} = 6 .5 \times 10^{- 7} C$
Distance between the spheres, r = 50 cm = 0.5 m
Force of repulsion between the two spheres, $F = \frac{1}{4 {πε}_{0}} \cdot \frac{q_{A} q_{B}}{r^{2}}$
$\begin{array}{l} \Rightarrow F = \frac{9 \times 10^{9} \times (6 .5 \times 10^{- 7})^{2}}{(0 .5)^{2}} \\ = 1 .52 \times 10^{- 2} N \end{array}$

(b)

When the charges are doubled and the distance is halved,
The force of repulsion between the two spheres,
$\begin{array}{l} F = \frac{1}{4 {πε}_{0}} \cdot \frac{2 q_{A} \times 2 q_{B}}{{(\frac{r}{2})}^{2}} = 16 \times \frac{q_{A} \times q_{B}}{r^{2}} \\ = 16 \times 1 .52 \times 10^{- 2} \\ = 0 .243 N \end{array}$

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