Question 281: Electrostats & Capacitance

Question

Two charges $7 \mu \mathrm{C}$ and $-2 \mu \mathrm{C}$ are placed at $(-9,0,0) \mathrm{cm}$ and $(9,0,0) \mathrm{cm}$ respectively in an external field $E=\frac{\mathrm{A}}{r^2} \hat{r}$, where $A=9 	imes 10^5 \mathrm{~N} / \mathrm{C} . \mathrm{m}^2$. Considering the potential at infinity is 0 , the electrostatic energy of the configuration is $\_\_\_\_$ J.

Accepted Answer

The total electrostatic energy $(\mathrm{U})$ of this system consists of two parts :
The energy of each individual charge due to the external electric field.
The interaction energy between the two charges.
The external electric field is given as $E=\frac{\mathrm{A}}{r^2}$.
The relation between electric field and potential ( V ) is:
$$ V(r)=-\int E \cdot d r=-\int \frac{A}{r^2} d r=\frac{A}{r} $$
The energy of a charge $q$ at distance $r$ is $U=q V(r)=\frac{q A}{r}$.
For $\mathrm{q}_1=7 \mu \mathrm{C}=7 	imes 10^{-6} \mathrm{C}$ :
Position $\mathrm{r}_1=9 \mathrm{~cm}=0.09 \mathrm{~m}$.
$$ U_1=\frac{q_1 A}{r_1}=\frac{\left(7 	imes 10^{-6}ight)\left(9 	imes 10^5ight)}{0.09}=\frac{6.3}{0.09}=70 \mathrm{~J} $$
For $\mathrm{q}_2=-2 \mu \mathrm{C}=-2 	imes 10^{-6} \mathrm{C}$ :
Position $\mathrm{r}_2=9 \mathrm{~cm}=0.09 \mathrm{~m}$.
$$ U_2=\frac{q_2 A}{r_2}=\frac{\left(-2 	imes 10^{-6}ight)\left(9 	imes 10^5ight)}{0.09}=-\frac{1.8}{0.09}=-20 \mathrm{~J} $$
The energy due to the interaction between the two point charges is given by :
$$ \mathrm{U}_{\mathrm{int}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{~d}} $$
The distance (d) between ( $-9,0,0$ ) and ( $9,0,0$ ) is $18 \mathrm{~cm}=0.18 \mathrm{~m}$.
Using $\mathrm{k}=9 	imes 10^9 \frac{\mathrm{~N} \cdot \mathrm{~m}^2}{\mathrm{C}^2}$ :
$$ U_{	ext {int }}=\frac{\left(9 	imes 10^9ight)\left(7 	imes 10^{-6}ight)\left(-2 	imes 10^{-6}ight)}{0.18} $$
$$\Rightarrow $$ $U_{\mathrm{int}}=-\frac{0.126}{0.18}=-0.7 \mathrm{~J}$
So, total electrostatic energy $\left(\mathrm{U}_{	ext {total }}ight)$,
$$ \mathrm{U}_{	ext {total }}=\mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_{\mathrm{int}} $$
$$\Rightarrow $$ $\mathrm{U}_{	ext {total }}=70+(-20)+(-0.7)$
$$\Rightarrow $$ $\mathrm{U}_{	ext {total }}=50-0.7=49.3 \mathrm{~J}$

Answer

49.3

Answer

1.4

Answer

24.3

Answer

-90.7

Electrostats & Capacitance

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