Electrostatics MCQs for NEET — Physics Questions with Answers

Force on each charge is zero. But if any of the charge is displaced, the net force starts acting on all of them.

Because electric field applies the force on electron in the direction opposite to it’s motion.

Energy density $=\frac{\text{Energy}}{\text{Volume}}$

so it’s dimensions are $\frac{M{L}^2{T}^{-2}}{L^3}=\left[M{L}^{-1}{T}^{-2}\right]$ 

Firstly being a conductor it is attracted by the high voltage plate, when charge is shared, ball is repelled until it goes to other plate and whole of the charge is transferred to the earth and the process is repeated.

For non-conducting sphere $E_{in}=\frac{k.Qr}{R^3}=\frac{\rho r}{3{\epsilon }_0}$ 

Net field at origin $E=\frac{q}{4\pi {\epsilon }_0}\left[\frac{{\displaystyle 1}}{{\displaystyle 1^2}}+\frac{{\displaystyle 1}}{{\displaystyle 2^2}}+\frac{{\displaystyle 1}}{{\displaystyle 4^2}}+....\infty \right]$

$=\frac{q}{4\pi {\epsilon }_0}\left[1+\frac{{\displaystyle 1}}{{\displaystyle 4}}+\frac{{\displaystyle 1}}{{\displaystyle 16}}+.....\infty \right]$

Using formula for infinite Geometric progression (G.P.) ,

if Sum = $a + \frac{a}{r} + \frac{a}{r^2} + \frac{a}{r^3} ........................\infty$

 and r<1 

then ,sum = $\frac{a}{1-r}$

In question, a =1 and r= $\frac{1}{4}$

$\therefore Sum = \frac{1}{1-{\displaystyle \frac{1}{4}}}=\frac{4}{3}$

$=\frac{q}{4\pi {\epsilon }_0}\left[\frac{{\displaystyle 1}}{{\displaystyle 1-\frac{{\displaystyle 1}}{{\displaystyle 4}}}}\right]=12\times {10}^{9}qN/C$  

In the given scenario of two infinitely long parallel conducting plates with opposite surface charge densities +σ and -σ separated by a small distance in vacuum, the electric field between the plates is given by E = σ/ε0, where ε0 is the permittivity of free space.

In electric dipole, the flux coming out from positive charge is equal to the flux coming in at negative charge i.e. total charge on sphere = 0. From Gauss law, total flux passing through the sphere = 0.

According to Gauss law $\oint E\cdot ds==\frac{ql}{{\epsilon }_0}$

$\oint ds=2\pi rl;$ (E is constant)

∴ $E\cdot 2\pi rl=\frac{ql}{{\epsilon }_0}$

⇒ $E=\frac{q}{2\pi {\epsilon }_{0}r}$ i.e. $E\propto \frac{1}{r}$

Two equal negative charges -q are fixed on the Y-axis, and a positive charge Q is released from rest on the X-axis. The net force on Q will be attractive towards the origin, but the force will not be along the line joining Q and the origin, causing oscillatory but not simple harmonic motion.