Question

(i) Determine electrostatic potential energy of a system consisting of two charges 7 μC and – 2μC (and with no external field) placed at (– 9 cm, 0, 0) and (9 cm, 0,0) respectively.
(ii) How much work is required to separate the two charges infinitely away from each other?
(iii) Suppose that the same system of charges is now placed in an external field
$E = A . \frac{1}{r^{2}};$ $A = 9 \times 10^{5} {cm}^{- 2}$ , What would the electrostatic energy of the configuration be?

Solution

Given,
Charge, q₁= $7 μC = 7 \times 10^{- 6} C$
Charge, q₂= $- 2 μC = - 2 \times 10^{- 6} C$
Distance betwen the charges, d=9-(-9)=18 cm= $18 \times 10^{- 2} m$

i) Electrostatic potential energy of the system is,
$U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r} = 9 \times 10^{9} \times \frac{(7 \times 10^{- 6}) \times (- 2 \times 10^{- 6})}{18 \times 10^{- 2}} = - 0.7 J$

(ii) Work done to seperate the two charges infinitely away from each other

$W = U_{2} - U_{1} = 0 - U = 0 - (- 0.7)$
= 0.7 J.

(c) Given,
$r_{1} = r_{2} = 9 cm = 0.09 m$

When, the system is placed in an external electric field, electrostatic energy is,

$U_{r} = q_{1} V (\vec{r_{1}}) + q_{2} V (\vec{r_{2}}) + \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r}$

$= q_{1} {Er}_{1} + q_{2} {Er}_{2} + \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r}$ [ E=V/r]

$= q_{1} \frac{A}{r_{1}^{2}} r_{1} + q_{2} . \frac{A}{r_{2}^{2}} . r_{2} + \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r}$
$= A \frac{q_{1}}{r_{1}} + A \frac{q_{2}}{r_{2}} + \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r} = 9 \times 10^{5} \times \frac{7 \times 10^{- 6}}{0.09} + 9 \times 10^{5} \times \frac{- 2 \times 10^{- 6}}{0.09} - 0.7 = 70 - 20 - 0.7 = 49.3 joule$

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