Question 355: Electrostats & Capacitance

Question

Two charges $7 \mu \mathrm{c}$ and $-4 \mu \mathrm{c}$ are placed at $(-7 \mathrm{~cm}, 0,0)$ and $(7 \mathrm{~cm}, 0,0)$ respectively. Given, $\epsilon_0=8.85 	imes 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the electrostatic potential energy of the charge configuration is :

Accepted Answer

$$ U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} $$
Here,
$$ q_1 = 7 	imes 10^{-6} \, C $$
$$ q_2 = -4 	imes 10^{-6} \, C $$
The separation between the charges is along the x-axis from $$ -7 \, 	ext{cm} $$ to $$ 7 \, 	ext{cm} $$, so
$$ r = 0.14 \, m $$
The Coulomb constant is approximated as
$$ \frac{1}{4\pi\epsilon_0} \approx 9 	imes 10^9 \, \frac{Nm^2}{C^2} $$
Substitute these values into the equation:
$$ U = 9 	imes 10^9 	imes \frac{(7 	imes 10^{-6})(-4 	imes 10^{-6})}{0.14} $$
Calculate the product of the charges:
$$ (7 	imes 10^{-6})(-4 	imes 10^{-6}) = -28 	imes 10^{-12} \, C^2 $$
Now, divide by the distance:
$$ \frac{-28 	imes 10^{-12}}{0.14} = -2 	imes 10^{-10} \, C^2/m $$
Finally, multiply by the Coulomb constant:
$$ U = 9 	imes 10^9 	imes (-2 	imes 10^{-10}) = -1.8 \, J $$
Thus, the electrostatic potential energy of the configuration is:
$$ -1.8 \, J $$
This corresponds to Option D.

Answer

$-2.0$ J

Answer

$-1.5$ J

Answer

$-1.2$ J

Answer

$-1.8$ J

Electrostats & Capacitance

Two charges 7 \mu \mathrm{c} and -4 \mu \mathrm{c} are placed at (-7 \mathrm{~cm}, 0,0) and (7 \mathrm{~cm}, 0,0) respectively. Given, \epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}, the electrostatic potential energy of the charge configuration is :

Choose the correct answer: