Question

# Two positive point charges are of $$12C$$ and $$8C$$ are $$10cm$$ apart from each other. The work done in bringing them $$4cm$$ closer is

A
5.8J
B
13eV
C
5.8eV
D
13J

Solution

## The correct option is D $$13J$$Given two charges $$q_{1}=12\mu C$$, $$q_{2}=8\mu C$$ are at distance '$$d_{1}$$'$$=10\, cm$$We have to find the work done in bringing them $$d_{2}=4\, cm$$ closer.For this, we find the charge in the potential energy of the system and then take the difference.Now $$d_{1}=10\, cm=10\times 10^{-2}=10^{-1}\, m$$$$d_{2}=4\, cm=4\times 10^{-2}\, m$$So, potential ($$V_{1}$$) when $$q_{1}$$ and $$q_{2}$$ are distance $$d_{1}$$ apart, $$V_{1}=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{d_{1}}$$Potential ($$V_{2}$$) when $$q_{1}$$ and $$q_{2}$$ are distance $$d_{2}$$ apart, $$V_{2}=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{d_{2}}$$Work done $$=V_{2}-V_{1}=\dfrac{1}{4\pi \varepsilon _{0}}q_{1}q_{2}\left ( \dfrac{1}{d_{2}}-\dfrac{1}{d_{1}} \right )$$$$=9\times 10^{9}\times 12\times 8\times 10^{-12}\left ( \dfrac{1}{4\times 10^{-2}}-\dfrac{1}{10\times 10^{-2}} \right )$$$$=864\times 10^{-3}\left [ \dfrac{6\times 10^{2}}{40\times 10^{-2}} \right ]$$$$=129.6\times 10^{1}$$$$=12.96\approx 13\; Joules$$Physics

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