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A positive charge .Q. is fixed at a poin...

A positive charge .Q. is fixed at a point, A negatively charged of mass .m. and charge .q. is revolving in a circular path of radius radius `r_(1)` with .Q. as the centre. The change in potential energy to change the radius of the circular path from `r_(1) and r_(2)` in joule is

A

zero

B

`(1)/(4piino)Qq[(1)/(r_(1))-(1)/(r_(2))]`

C

`(1)/(piino)Qq[(1)/(r_(1))-(1)/(r_(2))]`

D

`(Qq)/(4piino)[(1)/(r_(2))-(1)/(r_(1))]`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to calculate the change in potential energy when the radius of the circular path of a negatively charged particle changes from \( r_1 \) to \( r_2 \) while revolving around a fixed positive charge \( Q \). ### Step-by-Step Solution: 1. **Understanding the System**: - We have a positive charge \( Q \) fixed at a point. - A negatively charged particle with charge \( q \) and mass \( m \) is revolving in a circular path of radius \( r_1 \) around the charge \( Q \). 2. **Potential Energy Calculation**: - The potential energy \( U \) of a charge \( q \) in the electric field of another charge \( Q \) at a distance \( r \) is given by: \[ U = -\frac{k Q q}{r} \] - Here, \( k \) is Coulomb's constant, \( k = \frac{1}{4 \pi \epsilon_0} \). 3. **Potential Energy at Radius \( r_1 \)**: - The potential energy \( U_1 \) when the radius is \( r_1 \): \[ U_1 = -\frac{k Q q}{r_1} \] 4. **Potential Energy at Radius \( r_2 \)**: - The potential energy \( U_2 \) when the radius is \( r_2 \): \[ U_2 = -\frac{k Q q}{r_2} \] 5. **Change in Potential Energy**: - The change in potential energy \( \Delta U \) when the radius changes from \( r_1 \) to \( r_2 \) is given by: \[ \Delta U = U_2 - U_1 \] - Substituting the expressions for \( U_1 \) and \( U_2 \): \[ \Delta U = \left(-\frac{k Q q}{r_2}\right) - \left(-\frac{k Q q}{r_1}\right) \] - Simplifying this gives: \[ \Delta U = -\frac{k Q q}{r_2} + \frac{k Q q}{r_1} = k Q q \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \] 6. **Final Expression**: - Therefore, the change in potential energy when the radius changes from \( r_1 \) to \( r_2 \) is: \[ \Delta U = k Q q \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]
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