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

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

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

A and B are two concentric metallic shells. If A is positively charged and B is earthed, then electric

1.  Field at common centre is non-zero

2.  Field outside B is nonzero

3.  Potential outside B is positive

4.  Potential at common centre is positive