Question

Determine the electrostatic potential energy of a system consisting of two charges 7c and 2c (and with no external field) placed at (9cm,0,0) and (9cm,0,0) respectively.Mple 2.5 (a) determine the electrostatic potential energy of a system consisting of

Answer 1

Answer:

U=14π∈0q1q2r

=9×109×7×(−2)×10−120.18

=−0.7J

The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,

q1V(r1)+q2V(r2)=AμC0.09m+A−2μC−0.09m

and the net electrostatic energy is

=70−20−0.7=49.3J

Answer 2

$\huge{\underline{\underline{\tt{Solution:-}}}}$

(a)

$\longrightarrow \: \sf{U = \frac{1}{4 \pi \epsilon0} \frac{q _{1} \times q _{2}}{r} } \\ \\ \longrightarrow \: \sf{9 \times 10 {}^{9} \times \frac{7 \times ( - 2) \times 10 {}^{ - 12} }{0.18}} \\ \\ \implies \bf{ - 0.7 \: joules}$

(b) $\sf{W= U_{2}-U_{1}}$

$\longrightarrow \: \sf{0 - ( - 0.7)} \\ \\ \longrightarrow \: \bf{0.7 \: joules}$

(c) The mutual interaction energy of the two charges remains unchanged. In addition there is the energy of interaction of the two charges with the external field.

q1V(r1)+q2V(r2)

= A 7μC/0.09 m + A -2μC/0.09 m

and the net electrostatic energy is

q1V(r1)+q2V(r2)+q1q2/4πE0r12

= A7μC/0.09m + A-2μC/0.09 m-0.7 J

= 70-20-0.7

= 49.3 J