An elementary particle of mass m and charge e is projected with velocity v at a much more massive particle of charge Ze, where Z>0. What is the closest possible approach of the incident particle ?

1. $\frac{Z{e}^2}{2\pi {\epsilon }_{0}m{v}^2}$​

2. $\frac{Ze}{4\pi {\epsilon }_{0}m{v}^2}$

3. $\frac{Z{e}^2}{8\pi {\epsilon }_{0}m{v}^2}$​

4. $\frac{Ze}{8\pi {\epsilon }_{0}m{v}^2}$

Four equal charges Q are placed at the four corners of a square of each side is ‘a’. Work done in removing a charge – Q from its centre to infinity is 

(1) 0

(2) $\frac{\sqrt{2}{Q}^2}{4\pi {\epsilon }_{0}a}$

(3) $\frac{\sqrt{2}{Q}^2}{\pi {\epsilon }_{0}a}$

(4) $\frac{Q^2}{2\pi {\epsilon }_{0}a}$

An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed of the electron will be 

(1) $V\sqrt{e/m}$

(2) $\sqrt{eV/m}$

(3) $\sqrt{2eV/m}$

(4) $2eV/m$

Two equal charges q are placed at a distance of 2a and a third charge –2q is placed at the midpoint. The potential energy of the system is -

1. $\frac{q^2}{8\pi {\epsilon }_{0}a}$

2. $\frac{6q^2}{8\pi {\epsilon }_{0}a}$

3. $-\frac{7q^2}{8\pi {\epsilon }_{0}a}$

4. $\frac{9q^2}{8\pi {\epsilon }_{0}a}$