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.

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

q_{1}

and

q_{2}

, 1C each are at infinite distance from each other initially. They are then brought to 1m distance in vacuum. What was the initial potential energy of the system?

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 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.
$Δ P E = \frac{k Q q}{r f} - \frac{k Q q}{r i} = k Q q (\frac{1}{r f} - \frac{1}{r i}) = 9 \times 10^{9} (\frac{1}{5} - \frac{1}{10}) = 9 \times 10^{8} J$