There are two point charges of +1C each separated by 10m in vacuum. Calculate the change in potential energy of the system if they are brought to a separation of 5m.

$9 \times 10^{9} J$

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$9 \times 10^{8} J$

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$9 \times 10^{7} J$

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None

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Solution

The correct option is B
$9 \times 10^{8} J$

Change in potential energy will simply be final P.E minus initial P.E.
$Δ P E = \frac{k Q q}{r f} - \frac{k Q q}{r i} = k Q q (\frac{1}{r f} - \frac{1}{r i}) = 9 \times 10^{9} (\frac{1}{5} - \frac{1}{10}) = 9 \times 10^{8} J$

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Similar questions

There are two point charges of +1C each separated by 10m in vacuum. Calculate the change in potential energy of the system if they are brought to a separation of 5m.

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There are two point charges of +1C each separated by 10m in vacuum. Calculate the change in potential energy of the system if they are brought to a separation of 5m.

The correct option is B 9×108J Change in potential energy will simply be final P.E minus initial P.E. ΔPE=kQqrf−kQqri=kQq(1rf−1ri)=9×109(15−110)=9×108J

The correct option is B
$9 \times 10^{8} J$

Change in potential energy will simply be final P.E minus initial P.E.
$Δ P E = \frac{k Q q}{r f} - \frac{k Q q}{r i} = k Q q (\frac{1}{r f} - \frac{1}{r i}) = 9 \times 10^{9} (\frac{1}{5} - \frac{1}{10}) = 9 \times 10^{8} J$