ElectrostaticsHard

Question

Two charges $7\mu C$ and $- 2\mu C$ are placed at $( - 9,0,0)cm$ and $(9,0,0)cm$ respectively in an external field $E = \frac{A}{r^{2}}\widehat{r}$, where $A = 9 \times 10^{5}\text{ }N/C.m^{2}$.

Considering the potential at infinity is 0 , the electrostatic energy of the configuration is $\_\_\_\_$ J.

Options

A.1.4
B.-90.7
C.49.3
D.24.3

Solution

$\ \underset{( - 9,0,0)}{q_{1} = 7\mu C}\ r\ \underset{(0,0,0)}{\mapsto}\ r\ q_{2} = - 2\mu C$

$$\begin{matrix} & dV = - \overrightarrow{E}.\overrightarrow{dr} \\ & \ \int_{0}^{v}\mspace{2mu}\mspace{2mu} dV = - \int_{\infty}^{r}\mspace{2mu}\mspace{2mu}\frac{A}{r^{2}}dr \\ & V = - \left\lbrack \frac{- A}{r^{2}} \right\rbrack_{\infty}^{r} \Rightarrow V = \frac{A}{r} \\ & U = U_{\text{self~}} + U_{\text{interaction~}} \\ & \ = q_{1}v_{1} = q_{2}v_{2} + \frac{kq_{1}q_{2}}{2r} \\ & \ = 7 \times 10^{- 6}\frac{A}{9 \times 10^{- 2}} - 2 \times 10^{- 6}\frac{A}{9 \times 10^{- 2}} \\ & \ - \frac{9 \times 10^{9} \times 14 \times 10^{- 12}}{2 \times 9 \times 10^{- 2}} \\ & \ = \frac{5 \times 10^{- 6} \times 9 \times 10^{5}}{9 \times 10^{- 2}} - 7 \times 10^{- 1} \\ & \ = 50 - 0.7 \\ & \ = 49.3\text{ }J \end{matrix}$$

Create a free account to view solution

View Solution Free
Topic: Electrostatics·Practice all Electrostatics questions

More Electrostatics Questions