Question

Two charges-5 μC and +10 μC are placed 20 cm apart. The potential energy of the system is​

Answer 1

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Answer 2

Given:

Two charges-5 μC and +10 μC are placed 20 cm apart.

To find:

Potential energy of the system?

Calculation:

Let potential energy be PE :

$\rm \: PE = \dfrac{k(q1)(q2)}{d}$

• 'k' is Coulomb's Constant = 9 × 10⁹

Putting values in SI UNITS:

$\rm \implies\: PE = \dfrac{(9 \times {10}^{9}) ( - 5 \times {10}^{ - 6} )(10 \times {10}^{ - 6} )}{ \frac{20}{100} }$

$\rm \implies\: PE = \dfrac{(9 \times {10}^{9}) ( - 5 \times {10}^{ - 6} )( {10}^{ - 5} )}{ 0.2}$

$\rm \implies\: PE = \dfrac{ - 45 \times {10}^{ - 2}}{ 0.2}$

$\rm \implies\: PE = - 225\times {10}^{ - 2}$

$\rm \implies\: PE = - 2.25 \: joule$

So, potential energy of system is -2.25 Joule