A positively charged particle +q is projected with speed v toward a fixed charge +Q, and rebounds after reachi

Question

A positively charged particle +q is projected with speed v toward a fixed charge +Q, and rebounds after reaching a minimum distance r. What will be the new closest distance of approach if its initial velocity is doubled to 2v?

Accepted Answer

1. **Conservation of Energy:** According to the conservation of mechanical energy principles discussed in Class 11 Physics [2, 3], the initial kinetic energy of the particle is entirely converted into electric potential energy at the distance of closest approach (where the particle momentarily stops).
2. **Kinetic Energy Relation:** Kinetic energy ($K$) is given by $K = \frac{1}{2}mv^2$ [1]. Thus, $K \propto v^2$.
3. **Potential Energy Relation:** The electric potential energy ($U$) between two charges is inversely proportional to the distance ($r$) between them ($U \propto 1/r$), analogous to the gravitational potential energy discussed in the Class 11 Gravitation chapter [4].
4. **Derivation:** 
   - Initial state: $\frac{1}{2}mv^2 \propto \frac{1}{r} \implies r \propto \frac{1}{v^2}$.
   - New state: If the velocity is doubled ($v' = 2v$), the new kinetic energy becomes $K' \propto (2v)^2 = 4v^2$. This is 4 times the initial kinetic energy.
   - To balance this increased energy, the potential energy must increase by a factor of 4, which means the distance $r$ must decrease by a factor of 4.
   - $r' = \frac{r}{4}$.

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