Question

A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q was given a speed 2v, the closest distance of approach would be

• A. $ r $
• B. $ 2r $
• C. $ r/2 $
• D. $ r/4 $

Answer 1

Answer: (D)

Let a particle of charge q having velocity $ v $ approaches Q upto a closest distance r and if the velocity becomes $ 2v, $ the closest distance will be $ r $ . The law of conservation of energy yields, kinetic energy of particle = electric potential energy between them at closest distance of approach. Or $ \frac{1}{2}m{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{r} $ Or $ \frac{1}{2}m{{v}^{2}}=k\frac{Qq}{r} $ ...(i) $ \left( k=constant=\frac{1}{4\pi {{\varepsilon }_{0}}} \right) $ and $ \frac{1}{2}m{{(2v)}^{2}}=k\frac{Qq}{r} $ ?(ii) Dividing Eq. (i) by Eq. (ii), $ \frac{\frac{1}{2}m{{v}^{2}}}{\frac{1}{2}m{{(2v)}^{2}}}=\frac{\frac{kQq}{r}}{\frac{kQq}{r}} $ $ \Rightarrow $ $ \frac{1}{4}=\frac{r}{r} $ $ \Rightarrow $ $ r=\frac{r}{4} $