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

Two spheres of radius a and b respectively are charged and joined by a wire. The ratio of the electric field at the surface of the spheres is 

(1) a/b

(2) b/a

(3) a²/b²

(4) b²/a²

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) $2eV/m$

The dimension of (1/2) ${\epsilon }_0{E}^2 ({\epsilon }_0$: permittivity of free space; E: electric field) is

(1) MLT^–1

(2) ML²L^–2

(3) ML^–1T^–2

(4) ML²T^–1

A table tennis ball that has been covered with conducting paint is suspended by a silk thread so that it hangs between two plates, out of which one is earthed and other is connected to a high voltage generator. This ball

(1) Is attracted towards high voltage plate and stays there

(2) Hangs without moving

(3) Swing backward and forward hitting each plate in turn

(4) Is attracted to the earthed plate and stays there

Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle θ with the axis of the dipole, then the potential at P is given by (r >> 2a) (Where p = 2qa) 

(1) $V=\frac{p\cos \theta }{4\pi {\epsilon }_0{r}^2}$

(2) $V=\frac{p\cos \theta }{4\pi {\epsilon }_{0}r}$

(3) $V=\frac{p\sin \theta }{4\pi {\epsilon }_{0}r}$

(4) $V=\frac{p\cos \theta }{2\pi {\epsilon }_0{r}^2}$ 

A charge \(+q\) is fixed at each of the points $x=x_0,$ $x=3x_0,$ $x=5x_0$ ..... infinite, on the \(x\)-axis, and a charge \(-q\) is fixed at each of the points $x=2x_0,$ $x=4x_0,x=6x_0$,..... infinite. Here \(x_0\) is a positive constant. Take the electric potential at a point due to a charge \(Q\) at a distance \(r\) from it to be \(\frac{Q}{4\pi \varepsilon_0 r}\). Then, the potential at the origin due to the above system of charges is:
1. \(0\)
2. \(\frac{q}{8 \pi \varepsilon_{0} x_{0} \mathrm{ln} 2}\)
3. \(\infty\)
4. \(\frac{q \mathrm{ln} 2}{4 \pi \varepsilon_{0} x_{0}}\)

Point charge q moves from point P to point S along the path PQRS (figure shown) in a uniform electric field E pointing co-parallel to the positive direction of the x-axis. The coordinates of the points P, Q, R, and S are $(a,b,0),(2a,0,0),(a,-b,0)$ and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression 

(1) qEa

(2) – qEa

(3) $qEa\sqrt{2}$

(4) $qE\sqrt{[{\left(2a\right)}^2+b^2]}$

$\mathrm{The} \mathrm{electric} \mathrm{potential} \mathrm{at} \mathrm{a} \mathrm{point} \mathrm{at} \mathrm{distance} \sqrt{3}\mathrm{R} \mathrm{from} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{disc}$
$\mathrm{of} \mathrm{radius} \mathrm{R} \mathrm{lying} \mathrm{in} \mathrm{the} \mathrm{axis} \mathrm{of} \mathrm{the} \mathrm{disc} \mathrm{whose} \mathrm{surface} \mathrm{charge} \mathrm{density} \mathrm{is} \mathrm{\sigma }$
$\mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by}:$
$1. \frac{\sigma }{2{\epsilon }_0}\left[2-\sqrt{3}\right]R 2. \frac{\sigma }{2{\epsilon }_0}\left[2+\sqrt{3}\right]R$
$3.\frac{\sigma }{2{\epsilon }_0}\left[\sqrt{3}-\sqrt{2}\right]R 4. \frac{\sigma }{2{\epsilon }_0}\left[\sqrt{3}+\sqrt{2}\right]R$