Question

Two charges of 4uc and 7 uc are placed 4.5 cm apart .Find mutual P.E of the system​

Answer 1

Answer:

$\boxed{\mathfrak{Mutual \ P.E \ of \ the \ system = 4.8 \ J}}$

Given:

Value of two charges:

$\sf q_1 = 4 \: \mu C = 4 \times {10}^{ - 6} \: C$

$\sf q_2 = 7 \: \mu C = 7 \times {10}^{ - 6} \: C$

Separation between two charges:

$\sf r_{12} = 4.5 \: cm = 0.045 \: m$

We know:

$\sf k = 9 \times {10}^{9}$

To Find:

Mutual Potential Energy of the system (U)

Explanation:

Electric Potential Energy:

$\boxed{ \bold{\sf U = \frac{kq_1 q_2}{r_{12}}}}$

Substituting value of $\sf q_1, q_2, r_{12}$ & k in the equation:

$\sf \implies U = \frac{9 \times {10}^{9} \times 4 \times {10}^{ - 6} \times 7 \times {10}^{ - 6} }{0.045} \\ \\ \sf \implies U = \frac{9 \times 4 \times 6 \times {10}^{9 - 6 - 6} }{0.045} \\ \\ \sf \implies U = \frac{ \cancel{9 } \times 4 \times 6 \times {10}^{9 - (6 + 6)} }{ \cancel{9} \times 0.005} \\ \\ \sf \implies U = \frac{4 \times 6 \times {10}^{9 - 12} }{5 \times {10}^{ - 3} } \\ \\ \sf \implies U = \frac{4 \times 6 \times {10}^{ - 3} }{5 \times {10}^{ - 3} } \\ \\ \sf \implies U = \frac{24 \times {10 }^{ - 3 - ( - 3)} }{5} \\ \\ \sf \implies U = \frac{24 \times {10}^{ - 3 + 3} }{5} \\ \\ \sf \implies U = \frac{24 \times {10}^{0} }{5} \\ \\ \sf \implies U = \frac{24}{5} \\ \\ \sf \implies U =4.8 \: J$

$\therefore$

Mutual Potential Energy of the system (U) = 4.8 J