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Q.

Two like charges of magnitude $10^{- 9}$ coulomb each are separated by a distance of 2 m in free space. If the distance between the charges is increased to 4m then the ratio of electric potential energies is

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a

1 : 2

b

2 : 1

c

3 : 4

d

9 : 4

answer is B.

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Detailed Solution

q_{1} = 10^{- 9}; q_{2} = 10^{- 9}, r_{1} = 2 m, r_{2} = 4m

$U_{1} = \frac{1}{4 π \in_{0}} \times \frac{q_{1} q_{2}}{r_{1}} = 9 \times 10^{9} \times \frac{10^{- 9} \times 10^{- 9}}{2} = \frac{9 \times 10^{- 9}}{2}$

$U_{2} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r} = 9 \times 10^{9} \times \frac{10^{- 9} \times 10^{- 9}}{4} = \frac{9 \times 10^{- 9}}{4} \Rightarrow \frac{U_{1}}{U_{2}} = \frac{\frac{9 \times 10^{- 9}}{2}}{\frac{9 \times 10^{- 9}}{4}} = \frac{2}{1}$

$U_{1} : U_{2} = 2 : 1$

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