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 $\left( {{R_2} > {R_1}} \right)$

$$A.\dfrac{{Qq}}{2}\left[ {\dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_1}}}} \right] \\\ B.\dfrac{{Qq}}{2}\left[ {\dfrac{1}{{{R_2}}}} \right] \\\ C.KQq\left[ {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right] \\\ D. - KQq\left[ {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right] \\\ E.2KQq\left[ {\dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_1}}}} \right] \\\$$

Answer 1

In this question, we need to determine the work required to increase the radius of revolution of –q from ${R_1}$ to another orbit ${R_2}$. For this, we will follow Coulomb’s law, which is given as $F = K \cdot \dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$.

Complete step by step answer:

Following Coulomb’s law, the force of attraction or repulsion between the two charges of charge ${Q_1}$ and ${Q_2}$ with the distance between them as $r$ is directly proportional to $F\alpha \dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$. Removing the proportionality sign with the constant term, the force is given as $F = K \cdot \dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$ .

Here, ${Q_1} = Q,{\text{ }}{Q_2} = - q{\text{ and }}r = R$ so, substitute these values in the formula $F = K \cdot \dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$ to determine the expression of the force of attraction between the charges Q and –q.

$F = K \cdot \dfrac{{{Q_1}{Q_2}}}{{{r^2}}} \\\ = K \cdot \left( {\dfrac{{ - Qq}}{{{{\left( R \right)}^2}}}} \right) - - - - (i) \\\$

Now, the work done in moving a positive unit charge from the point in infinity to a point ‘a’ is given by the product of the force applied on the charge and the displacement of the charge. In other words, the product of the force and the displacement due to the force is known as work done. Mathematically $dW = F \cdot dR$. The net work done will be calculated by integrating over the bounded limits.

Here, the radius of the revolution should be increased from ${R_1}$ to another orbit ${R_2}$ , and so, the limits are from ${R_1}$ to ${R_2}$.

Integrating the equation (i) in the bounded limits we get,

$$\int {dW} = \int\limits_{{R_1}}^{{R_2}} {F \cdot dR} \\\ W = \int\limits_{{R_1}}^{{R_2}} {K \cdot \left( {\dfrac{{ - Qq}}{{{{\left( R \right)}^2}}}} \right)} dR \\\ = - KQq\int\limits_{{R_1}}^{{R_2}} {\left( {\dfrac{1}{{{{\left( R \right)}^2}}}} \right)} dR \\\ = - KQq\left[ {\dfrac{{ - 1}}{R}} \right]_{{R_1}}^{{R_2}} \\\ = - KQq\left[ {\dfrac{{ - 1}}{{{R_2}}} + \dfrac{1}{{{R_1}}}} \right] \\\ = KQq\left[ {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right] \\\$$

Hence, the work required to increase the radius of revolution of –q from ${R_1}$ to another orbit ${R_2}$ is $KQq\left[ {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right]$.

Option C is correct.

Note: Students should be careful while selecting the limits of the bounded region as here, the revolution of the charge –q is increased from ${R_1}$ to another orbit ${R_2}$ and so as the limits. Moreover, the opposite charges are involved here in this question, so; they attract each other.