What will be the electrical potential energy of a charge 7 NC that is 2 cm from a 20 nC charge
Answers
Explanation:
General expression for Electrostatic Potential Energy is:
\boxed{ \rm PE = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{(q1)(q2)}{d} \bigg \}}
PE=
4πϵ
0
1
{
d
(q1)(q2)
}
\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{(20 \times {10}^{ - 9} )(7 \times {10}^{ - 9} )}{ \frac{2}{100} } \bigg \}⟹PE=9×10
9
×{
100
2
(20×10
−9
)(7×10
−9
)
}
\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{(20 \times {10}^{ - 9} )(7 \times {10}^{ - 9} )}{ 2 \times {10}^{ - 2} } \bigg \}⟹PE=9×10
9
×{
2×10
−2
(20×10
−9
)(7×10
−9
)
}
\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{(2 \times {10}^{ - 8} )(7 \times {10}^{ - 9} )}{ 2 \times {10}^{ - 2} } \bigg \}⟹PE=9×10
9
×{
2×10
−2
(2×10
−8
)(7×10
−9
)
}
\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{( {10}^{ - 8} )(7 \times {10}^{ - 9} )}{ {10}^{ - 2} } \bigg \}⟹PE=9×10
9
×{
10
−2
(10
−8
)(7×10
−9
)
}
\implies\rm PE = 9 \times {10}^{9} \times \bigg \{ \dfrac{7 \times {10}^{ - 17} }{ {10}^{ - 2} } \bigg \}⟹PE=9×10
9
×{
10
−2
7×10
−17
}
\implies\rm PE = 9 \times {10}^{9} \times \bigg \{7 \times {10}^{ - 15} \bigg \}⟹PE=9×10
9
×{7×10
−15
}
\implies\rm PE = 63\times {10}^{ - 6} \: joule⟹PE=63×10
−6
joule
So, final answer is:
\boxed{ \bold{ PE = 63\times {10}^{ - 6} \: joule}}
PE=63×10
−6
joule