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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Solution

The correct option is B
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.

Assuming potential energy at infinite separation is zero, what is the Potential Energy of a two charge system if the charges are both +10C and are separated by 10m?

Q. Charges

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and

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Assuming potential energy at infinite separation is zero, what is the Potential Energy of a two charge system if the charges are both +10C and are separated by 10m?

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