Question

two point charges of +10 are placed at a distance 40 cm in air potential energy of the sistem will be

Answer 1

potential energy is 5.625×10^-2

Answer 2

The potential energy of the system will be $2.25\times 10^{12}\ J$.

Explanation:

It is given that,

Charges, $q_1=q_2=10\ C$

Distance between charges, d = 40 cm = 0.4 m

Let U is the potential energy of the system. The formula of the potential energy of the system is given by :

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

$U=\dfrac{9\times 10^9\times 10^2}{0.4}$

$U=2.25\times 10^{12}\ J$

So, the potential energy of the system will be $2.25\times 10^{12}\ J$. Hence, this is the required solution.

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