Two charges q_1 and q_2 are fixed at points A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The potential energy $U$ of a system of charges is given by $U = \sum \frac{1}{4\pi\epsilon_0} \frac{q_i q_j}{r_{ij}}$.Since $q_1$ and $q_2$ are f

Vedclass · Accepted Answer

$8q_2$

Vedclass · Answer

(A) The potential energy $U$ of a system of charges is given by $U = \sum \frac{1}{4\pi\epsilon_0} \frac{q_i q_j}{r_{ij}}$.Since $q_1$ and $q_2$ are fixed, the change in potential energy $\Delta U$ depends only on the change in the interaction energy of $q_3$ with $q_1$ and $q_2$.At point $C$: $r_{1C} = 40 	ext{ cm} = 0.4 	ext{ m}$, $r_{2C} = \sqrt{AC^2 + AB^2} = \sqrt{40^2 + 30^2} = 50 	ext{ cm} = 0.5 	ext{ m}$.Potential at $C$ due to $q_1, q_2$ is $V_C = \frac{1}{4\pi\epsilon_0} (\frac{q_1}{0.4} + \frac{q_2}{0.5})$.At point $D$: $r_{1D} = 40 	ext{ cm} = 0.4 	ext{ m}$, $r_{2D} = AD - AB = 40 - 30 = 10 	ext{ cm} = 0.1 	ext{ m}$.Potential at $D$ due to $q_1, q_2$ is $V_D = \frac{1}{4\pi\epsilon_0} (\frac{q_1}{0.4} + \frac{q_2}{0.1})$.Change in potential energy $\Delta U = q_3(V_D - V_C) = \frac{q_3}{4\pi\epsilon_0} [(\frac{q_1}{0.4} + \frac{q_2}{0.1}) - (\frac{q_1}{0.4} + \frac{q_2}{0.5})]$.$\Delta U = \frac{q_3}{4\pi\epsilon_0} [\frac{q_2}{0.1} - \frac{q_2}{0.5}] = \frac{q_3}{4\pi\epsilon_0} [10q_2 - 2q_2] = \frac{q_3}{4\pi\epsilon_0} [8q_2]$.Comparing with $\frac{q_3 K}{4\pi\epsilon_0}$, we get $K = 8q_2$.

Similar Questions

The potential of a large liquid drop formed when eight identical small liquid drops are combined is $20 \,V$. What was the potential of each single small drop (in $\,V$)?

Statement-$1$: For a charged particle moving from point $P$ to point $Q$,the net work done by the electrostatic force is independent of the path connecting point $P$ to point $Q$.
Statement-$2$: The net work done by a conservative force on an object in a closed loop is zero.

Consider three concentric metal shells $A$,$B$,and $C$ with radii $a$,$b$,and $c$ respectively,as shown in the figure $(a < b < c)$. Their surface charge densities are $\sigma$,$-\sigma$,and $\sigma$ respectively. Calculate the electric potential on the surface of shell $A$.

Assertion: Electrons move away from a region of higher potential to a region of lower potential.
Reason: An electron has a negative charge.

$A$ charge of $10 \,\mu C$ is placed at the origin of an $x-y$ coordinate system. The potential difference between two points $(0, a)$ and $(a, 0)$ in volts will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module