Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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?

r/4

r/2

r/16

r/8

Step-by-Step Solution

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).
Kinetic Energy Relation: Kinetic energy ( $K$ ) is given by $K = \frac{1}{2}mv^2$ [1]. Thus, $K \propto v^2$ .
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].
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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