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Question

Two positive charges of $$20\mu C$$ and $$8\mu C$$ are 20 cm apart. Find the work done in bringing them 5cm closer.


Solution

$$q_1 = 20 \times 10^{-6} c \space , \space q_2 = 8 \times 10^{-6}c$$

So, the primary distance $$r_1 = 20cm = 0.2m$$
Final distance (20-5) = 15 = 0.15m

Now the work done = $$\dfrac{kq_1q_2}{R}$$
$$=\dfrac{9 \times 10^9 \times 20 \times 10^{-6} \times 8 \times 10^{-6}}{(0.15-0.2)} \times 0.2$$
$$= 5.7 Joule$$

Physics
NCERT
Standard XII

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