Question

Two point charges of 10micro coloumb and -10 micro coloumb are placed at a distance of 40 cm in air potential energy of the system will be

Answer 1

2.25 J..............

Answer 2

The potential energy of the system will be 2.25 Joules.

Explanation:

It is given that,

Charge 1, $q_1=10\ \mu C=10^{-5}\ C$

Charge 1, $q_1=-10\ \mu C=-10^{-5}\ C$

Distance between charges, d = 40 cm = 0.4 m

We need to find the potential energy of the system. It is defined as the product of electric potential and the other charge such that,

$U=\dfrac{kq_1q_2}{r}$

$U=\dfrac{9\times 10^9\times (10^{-5})^2}{0.4}$

U = 2.25 Joules

So, the potential energy of the system will be 2.25 Joules. Hence, this is the required solution.

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