Two point charges 4q and -q are fixed on the | vedclass

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

Question

(C) The potential energy of a charge $q$ at a point is given by $U = qV$,where $V$ is the electric potential due to the fixed charges.Initial position

Vedclass · Accepted Answer

decrease by $\frac{4q^2}{3\pi\varepsilon_0 d}$

Vedclass · Answer

(C) The potential energy of a charge $q$ at a point is given by $U = qV$,where $V$ is the electric potential due to the fixed charges.Initial position: $x_i = 0$. The distances from $4q$ and $-q$ are $r_1 = d/2$ and $r_2 = d/2$.Initial potential $V_i = \frac{1}{4\pi\varepsilon_0} \left( \frac{4q}{d/2} + \frac{-q}{d/2} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{3q}{d/2} \right) = \frac{6q}{4\pi\varepsilon_0 d}$.Final position: $x_f = d$. The distances from $4q$ and $-q$ are $r_1' = d + d/2 = 3d/2$ and $r_2' = d - d/2 = d/2$.Final potential $V_f = \frac{1}{4\pi\varepsilon_0} \left( \frac{4q}{3d/2} + \frac{-q}{d/2} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{8q}{3d} - \frac{2q}{d} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{8q - 6q}{3d} \right) = \frac{2q}{12\pi\varepsilon_0 d} = \frac{q}{6\pi\varepsilon_0 d}$.Change in potential energy $\Delta U = q(V_f - V_i) = q \left( \frac{q}{6\pi\varepsilon_0 d} - \frac{6q}{4\pi\varepsilon_0 d} \right) = \frac{q^2}{4\pi\varepsilon_0 d} \left( \frac{2}{3} - 6 \right) = \frac{q^2}{4\pi\varepsilon_0 d} \left( -\frac{16}{3} \right) = -\frac{4q^2}{3\pi\varepsilon_0 d}$.The negative sign indicates a decrease in potential energy by $\frac{4q^2}{3\pi\varepsilon_0 d}$.

Two point charges $4q$ and $-q$ are fixed on the $x$-axis at $x = -d/2$ and $x = d/2$,respectively. If a third point charge $q$ is moved from the origin to $x = d$ along the semicircle as shown in the figure,the potential energy of the charge will:

Similar Questions

Define electron volt and convert it into Joule unit.

Three charges,each $+q$,are placed at the corners of an isosceles triangle $ABC$ with sides $BC = AC = 2a$. $D$ and $E$ are the midpoints of $BC$ and $AC$,respectively. The work done in taking a charge $Q$ from $D$ to $E$ is

Eight equal charges $q$ are placed at each corner of a cube of side $a$. The work done in carrying a charge $-q$ from its centre to infinity is:

An electric charge $10^{-3} \, \mu C$ is placed at the origin $(0, 0)$ of an $X-Y$ coordinate system. Two points $A$ and $B$ are situated at $(\sqrt{2}, \sqrt{2})$ and $(2, 0)$ respectively. The potential difference between the points $A$ and $B$ will be $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module