Question 811: Electrostats & Capacitance

Question

Two point charges $\mathrm{q}_1=6 \mu \mathrm{C}$ and $\mathrm{q}_2=4 \mu \mathrm{C}$ are kept at points $A$ and $B$ in air where $A B=10 \mathrm{~cm}$. What is the increase in potential energy of the system when $\mathrm{q}_2$ is moved towards $\mathrm{q}_1$ by 2 cm ?
$$\left(\frac{1}{4 \pi \varepsilon_0}=9 	imes 10^9 	ext { SI units }ight)$$

Accepted Answer

The potential energy between two charges is defined by the formula:
$  U = \frac{K q_1 q_2}{r}  $
Calculating the Increase in Potential Energy:
Initial Potential Energy ($ U_i $)
:
The initial distance between the charges is $ d = 10 \, 	ext{cm} = 0.1 \, 	ext{m} $.
Initial potential energy $ U_i = \frac{K q_1 q_2}{d} $
Final Potential Energy ($ U_f $)
:
The charge $ q_2 $ is moved towards $ q_1 $ by $ x = 2 \, 	ext{cm} = 0.02 \, 	ext{m} $.
The new separation becomes $ d - x = 0.1 \, 	ext{m} - 0.02 \, 	ext{m} = 0.08 \, 	ext{m} $.
Final potential energy $ U_f = \frac{K q_1 q_2}{d - x} $
Increase in Potential Energy ($ \Delta U $)
:
The increase in potential energy is given by the difference:
$    \Delta U = U_f - U_i = K q_1 q_2 \left( \frac{1}{d - x} - \frac{1}{d} ight)    $
$    \Delta U = \frac{K q_1 q_2 x}{d(d - x)}    $
Substituting Given Values:
$ K = 9 	imes 10^9 \, 	ext{N m}^2/	ext{C}^2 $
$ q_1 = 6 \, \mu	ext{C} = 6 	imes 10^{-6} \, 	ext{C} $
$ q_2 = 4 \, \mu	ext{C} = 4 	imes 10^{-6} \, 	ext{C} $
$ x = 0.02 \, 	ext{m} $
$ d = 0.1 \, 	ext{m} $
Performing the calculation:
$ \Delta U = \frac{(9 	imes 10^9) 	imes (6 	imes 10^{-6}) 	imes (4 	imes 10^{-6}) 	imes 0.02}{0.1 	imes 0.08} $
Simplifying:
$ \Delta U = \frac{4.32 	imes 10^{-3}}{0.008} = 0.54 \, 	ext{J} $
Therefore, the increase in potential energy of the system is $ 0.54 \, 	ext{J} $.

Answer

0.27 J

Answer

0.54 J

Answer

0.81 J

Answer

54 J

Electrostats & Capacitance

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