Question

Two point charges 4 uC and 10 °C are placed at 20 cm distance. The electrostatic potential energy of the system is
0 3.6 J
@ 2.4 J
O 1.8 J
o 18 J​

Answer 1

Answer:U = 1.8 J

Explanation:

U = kQq/r

U = 9 * 10^9 * 10 * 4 * 10^-6/20 * 10^-2

U = 36 * 10^4/2 * 10^-2

U = 1.8 J

Answer 2

POTENTIAL ENERGY OF SYSTEM IS 1.8 J

Given:

• Charges 4uC and 10uC are placed 20 cm from each other.

To find:

Electrostatic potential energy of system ?

Calculation:

The general expression of electrostatic potential energy of a system of charges is :

$U = \dfrac{k q_{1} q_{2} }{d}$

• 'k' is COULOMB'S CONSTANT, q1 and q2 are the charges and d is separation.

Putting the values in SI UNIT:

$\implies U = \dfrac{(9 \times {10}^{9})(4 \times {10}^{ - 6} \times 10 \times {10}^{ - 6}) }{ \dfrac{20}{100} }$

$\implies U = \dfrac{(9 \times {10}^{9})(4 \times {10}^{ - 11}) }{2 \times {10}^{ - 1} }$

$\implies U = (9 \times {10}^{9})(2 \times {10}^{ - 10})$

$\implies U = 1.8 \: joule$

So, the net potential energy of system is 1.8 Joule