Question

what is electric potential energy of a 7nC charge that is 2cm from a 20nC charge?​

Answer 1

To find:

Electric potential energy of a 7nC charge that is 2cm from a 20nC charge?

Calculation:

General expression for Electrostatic Potential Energy is:

$\boxed{ \rm PE = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{(q1)(q2)}{d} \bigg \}}$

$\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{(20 \times {10}^{ - 9} )(7 \times {10}^{ - 9} )}{ \frac{2}{100} } \bigg \}$

$\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{(20 \times {10}^{ - 9} )(7 \times {10}^{ - 9} )}{ 2 \times {10}^{ - 2} } \bigg \}$

$\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{(2 \times {10}^{ - 8} )(7 \times {10}^{ - 9} )}{ 2 \times {10}^{ - 2} } \bigg \}$

$\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{( {10}^{ - 8} )(7 \times {10}^{ - 9} )}{ {10}^{ - 2} } \bigg \}$

$\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{7 \times {10}^{ - 17} }{ {10}^{ - 2} } \bigg \}$

$\implies\rm PE = 9 \times {10}^{9} \times \bigg \{7 \times {10}^{ - 15} \bigg \}$

$\implies\rm PE = 63\times {10}^{ - 6} \: joule$

So, final answer is:

$\boxed{ \bold{ PE = 63\times {10}^{ - 6} \: joule}}$

Answer 2

Answer:

What will be the electrical potential energy of a charge 7 NC that is 2 cm from a 20 nC charge