Two charges 7 ?C and - 2 ?C are placed at (?9,0,0) cm and (9,0,0) cm respectively in an external field E = A r 2 r , w h e r e A = 9 × 10 5 ? N/C · m 2 . Considering the potential at infinity is 0, the electrostatic energy of the configuration is

Question

Two charges $7 μ C and - 2 μ C$ are placed at (−9,0,0) cm and (9,0,0) cm respectively in an external field $\vec{E} = \frac{A}{r^{2}} \hat{r}, where A = 9 \times 10^{5} N/C \cdot m^{2}$ . Considering the potential at infinity is 0, the electrostatic energy of the configuration is ______ J. [2026]

Smart Achievers · Accepted Answer

(2)

$d V = - \vec{E} \cdot d \vec{r}$

$\int_{0}^{V} d V = - \int_{\infty}^{r} \frac{A}{r^{2}} d r$

$V = - {[\frac{- A}{r}]}_{\infty}^{r} \Rightarrow V = \frac{A}{r}$

$U = U_{self} + U_{interaction}$

$= q_{1} V_{1} = q_{2} V_{2} + \frac{k q_{1} q_{2}}{2 r}$

$= 7 \times 10^{- 6} \cdot \frac{A}{9 \times 10^{- 2}} - 2 \times 10^{- 6} \cdot \frac{A}{9 \times 10^{- 2}}$ $- \frac{9 \times 10^{9} \times 14 \times 10^{- 12}}{2 \times 9 \times 10^{- 2}}$

$= \frac{5 \times 10^{- 6} \times 9 \times 10^{5}}{9 \times 10^{- 2}} - 7 \times 10^{- 1}$

$= 50 0.7$

$= 49.3 J$

Solution

1 –90.7

2 49.3

3 1.4

4 24.3

Want Us to Email you About Special Offers & Updates?