Question 196: Electrostatics & Capacitance

Question

Two charges $q_1$ and $q_2$ are separated by a distance of 30 cm . A third charge $q_3$ initially at ' C ' as shown in the figure, is moved along the circular path of radius 40 cm from C to D . If the difference in potential energy due to movement of $q_3$ from C to D is given by $\frac{q_3 \mathrm{~K}}{4 \pi \epsilon_0}$, the value of K is :

Accepted Answer

To determine the value of $ K $ for the given electrical potential scenario, we start by calculating the potential at points C and D.
Potential at Point C:
The potential at point C, $ V_C $, is given by the sum of the potentials due to charges $ q_1 $ and $ q_2 $:
$ V_C = \frac{kq_1}{0.4} + \frac{kq_2}{0.5} $
Potential at Point D:
Similarly, the potential at point D, $ V_D $, is:
$ V_D = \frac{kq_1}{0.4} + \frac{kq_2}{0.1} $
Change in Potential Energy:
The change in potential energy $ \Delta U $ when $ q_3 $ moves from C to D is:
$ \Delta U = (V_D - V_C) 	imes q_3 $
Substituting the expressions for $ V_D $ and $ V_C $, we get:
$ \Delta U = \left(\frac{kq_2}{0.1} - \frac{kq_2}{0.5}ight) 	imes q_3 $
Simplifying the Expression:
By simplifying the expression, we find:
$ \Delta U = \left(\frac{5kq_2}{0.5}ight) 	imes q_3 - \left(\frac{kq_2}{0.5}ight) 	imes q_3 $
$ \Delta U = \frac{4kq_2}{0.5} 	imes q_3 = 8kq_2 q_3 $
$ \Delta U = \frac{8q_2 q_3}{4\pi \varepsilon_0} $
Thus, the value of $ K $ is $ 8q_2 $.

Answer

$6 \mathrm{q}_2$

Answer

$6 \mathrm{q}_1$

Answer

$8 \mathrm{q}_1$

Answer

$\mathrm{8 q_2}$

Electrostatics & Capacitance

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