Question

Two charged particles, with equal charges of 2.0 × 10−5 C, are brought from infinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.

Answer 1

36 J Increase in the production of electric potential energy.

Explanation:

Charging magnitude, q1 = q2 = 2 X 10−5 C  

Initial potential energy = 0  

Each is transferred from infinity to within 10 cm.  

In other words, r = 10 = 10 X $10^{-2}$ m  

So work done = negative of work done. (Potential E)  

That is, the potential for increased energy,

$\begin{equation}\begin{aligned}&P . E=\int_{\infty}^{10} F \times d S\\&| P . E=K \frac{q_{1} q_{2}}{r}\end{aligned}\end$

K=9 X $10^9$

$\begin{equation}\begin{aligned}&P . E=\left(9 \times 10^{9}\right) \frac{2 \times 10^{-5} \times 2 \times 10^{-5}}{10 \times 10^{-2}}\\&P \cdot E=\frac{9 \times 10^{9} \times 4 \times 10^{-10}}{10 \times 10^{-2}}\\&P \cdot E=\frac{36 \times 10^{9} \times 10^{-10}}{10 \times 10^{-2}}\\&\begin{aligned}P . E &=\frac{36 \times 10^{-1}}{10 \times 10^{-2}} \\P . E &=\frac{36 \times 10}{10} \\P . E &=36 \mathrm{J}\end{aligned}\end{aligned}\end$

36 J Increase in the production of electric potential energy.

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

36 Joule increase in electric potential