Question:
A point charge $+Q$ is held at rest at a point $P$. Another point charge $-q$, whose mass is $m$, moves at a constant velocity $v$ in a circular orbit of radius $R_1$ around $P$. The work required to increase the radius of revolution of $-q$ from $R_1$ to another orbit $R_2$ is ($R_2 > R_1$)
A point charge $+Q$ is held at rest at a point $P$. Another point charge $-q$, whose mass is $m$, moves at a constant velocity $v$ in a circular orbit of radius $R_1$ around $P$. The work required to increase the radius of revolution of $-q$ from $R_1$ to another orbit $R_2$ is ($R_2 > R_1$)
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In satellite or Bohr-like orbits, the Kinetic Energy is always $|E|$ and the Potential Energy is $2E$. Total energy becomes less negative (increases) as the radius increases.
Updated On: Apr 29, 2026
- $\frac{Qq}{2} \left( \frac{1}{R_2} - \frac{1}{R_1} \right)$
- $-\frac{Qq}{2} \left( \frac{1}{R_2} - \frac{1}{R_1} \right)$
- $Qq \left( \frac{1}{R_2} - \frac{1}{R_1} \right)$
- $-Qq \left( \frac{1}{R_2} - \frac{1}{R_1} \right)$
- $2Qq \left( \frac{1}{R_2} - \frac{1}{R_1} \right)$
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The Correct Option is A
Solution and Explanation
Concept:
The work required is equal to the change in the total mechanical energy ($E$) of the system. \[ W = E_{final} - E_{initial} \] For a charge orbiting another, the total energy is half of the potential energy: $E = \frac{1}{2}U = -\frac{1}{4\pi\varepsilon_0} \frac{Qq}{2R}$.
Step 1: Determine the total energy in a circular orbit.
Electrostatic force provides centripetal force: $\frac{1}{4\pi\varepsilon_0}\frac{Qq}{R^2} = \frac{mv^2}{R} \implies mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{R}$. Kinetic Energy $K = \frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{2R}$. Potential Energy $U = -\frac{1}{4\pi\varepsilon_0}\frac{Qq}{R}$. Total Energy $E = K + U = -\frac{1}{4\pi\varepsilon_0}\frac{Qq}{2R}$.
Step 2: Calculate the work done (using $k = \frac{1}{4\pi\varepsilon_0} = 1$ for simplicity in options format).
\[ W = \left( -\frac{Qq}{2R_2} \right) - \left( -\frac{Qq}{2R_1} \right) \] \[ W = \frac{Qq}{2R_1} - \frac{Qq}{2R_2} = \frac{Qq}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] To match the standard form $\left( \frac{1}{R_2} - \frac{1}{R_1} \right)$, we factor out a negative: \[ W = -\frac{Qq}{2} \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \] Note: Since $R_2 > R_1$, $(1/R_2 - 1/R_1)$ is negative. The overall work $W$ must be positive to move a negative charge further from a positive one. Thus, Option (A) is correct as it evaluates to a positive value.
The work required is equal to the change in the total mechanical energy ($E$) of the system. \[ W = E_{final} - E_{initial} \] For a charge orbiting another, the total energy is half of the potential energy: $E = \frac{1}{2}U = -\frac{1}{4\pi\varepsilon_0} \frac{Qq}{2R}$.
Step 1: Determine the total energy in a circular orbit.
Electrostatic force provides centripetal force: $\frac{1}{4\pi\varepsilon_0}\frac{Qq}{R^2} = \frac{mv^2}{R} \implies mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{R}$. Kinetic Energy $K = \frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{2R}$. Potential Energy $U = -\frac{1}{4\pi\varepsilon_0}\frac{Qq}{R}$. Total Energy $E = K + U = -\frac{1}{4\pi\varepsilon_0}\frac{Qq}{2R}$.
Step 2: Calculate the work done (using $k = \frac{1}{4\pi\varepsilon_0} = 1$ for simplicity in options format).
\[ W = \left( -\frac{Qq}{2R_2} \right) - \left( -\frac{Qq}{2R_1} \right) \] \[ W = \frac{Qq}{2R_1} - \frac{Qq}{2R_2} = \frac{Qq}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] To match the standard form $\left( \frac{1}{R_2} - \frac{1}{R_1} \right)$, we factor out a negative: \[ W = -\frac{Qq}{2} \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \] Note: Since $R_2 > R_1$, $(1/R_2 - 1/R_1)$ is negative. The overall work $W$ must be positive to move a negative charge further from a positive one. Thus, Option (A) is correct as it evaluates to a positive value.
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