Question 1:
easy1. The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \AA$ apart is.
($m_e \simeq 9 \times 10^{-31} \text{ kg}, e = 1.6 \times 10^{-19} \text{ C}$)
(Take $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}$)
(2020-Covid)
Using Coulomb's law, $F = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$. Acceleration $a = \frac{F}{m_e}$. Plugging in values gives $a = 10^{22} \text{ m/s}^2$.
Question 2:
easy2. Suppose the charge of a proton and an electron differ slightly. One of them is $-e$, the other is $(e + \Delta e)$. If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance $d$ (much greater than atomic size) apart is zero, then $\Delta e$ is of the order of [Given mass of hydrogen $m_h = 1.67 \times 10^{-27} \text{ kg}$] (2017-Delhi)
Net charge on each atom is $\Delta e$. Equating electrostatic repulsion to gravitational attraction: $\frac{1}{4\pi\epsilon_0} \frac{(\Delta e)^2}{d^2} = \frac{G m_h^2}{d^2}$. Solving yields $\Delta e \approx 10^{-37} \text{ C}$.
Question 3:
easy3. Two identical charged spheres suspended from a common point by two massless strings of lengths $\ell$, are initially at a distance $d$ ($d \ll \ell$) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $V$. Then $V$ varies as a function of the distance $x$ between the spheres, as: (2016 – I)
For equilibrium, $\tan\theta = \frac{F_e}{mg} \implies \frac{x}{2\ell} = \frac{k q^2}{x^2 mg} \implies q \propto x^{3/2}$. Differentiating w.r.t time, $\frac{dq}{dt} \propto x^{1/2} V$. Since $\frac{dq}{dt}$ is constant, $V \propto x^{-1/2}$.
Question 4:
easy5. Two positive ions, each carrying a charge $q$, are separated by a distance $d$. If $F$ is the force of repulsion between the ions, the number of electrons missing from each ion will be ($e$ being the charge on an electron): (2010 Pre)
Coulomb force $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2}$, so $q = \sqrt{4\pi\epsilon_0 F d^2}$. Number of missing electrons is $n = \frac{q}{e} = \sqrt{\frac{4\pi\epsilon_0 F d^2}{e^2}}$.
Question 5:
easy6. The unit of permittivity of free space $\epsilon_0$ is: (2004)
From Coulomb's Law, $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$. Rearranging gives $\epsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$. The unit is $\text{Coulomb}^2/\text{Newton metre}^2$.
Question 6:
easy7. An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius $r$. The Coulomb force $\vec{F}$ between the two is: (2003)
(where $K = \frac{1}{4\pi\epsilon_0}$)
The Coulomb force is attractive, so $\vec{F} = -K \frac{e^2}{r^2} \hat{r}$. Multiplying numerator and denominator by $r$ (since $\hat{r} = \frac{\vec{r}}{r}$) gives $\vec{F} = -K \frac{e^2}{r^3} \vec{r}$.
Question 7:
easy8. A charge $q$ is placed at the centre of the line joining two exactly equal positive charges $Q$. The system of three charges will be in equilibrium, if $q$ is equal to (1995)
For the system to be in equilibrium, the net force on any charge must be zero. Considering a charge $Q$ at the end, $F_{\text{net}} = \frac{k Q^2}{x^2} + \frac{k Q q}{(x/2)^2} = 0$. Solving this yields $q = -Q/4$.
Question 8:
easy9. Point charges $+4q$, $-q$ and $+4q$ are kept on the X-axis at point $x = 0$, $x = a$ and $x = 2a$ respectively. (1988)
The net force on each charge is zero, so all are in equilibrium. However, displacing any charge slightly along the axis results in a net force acting in the direction of displacement, meaning the equilibrium is unstable for all (Earnshaw's theorem).
Question 9:
easyA spherical conductor of radius $10 \text{ cm}$ has a charge of $3.2 \times 10^{-7} \text{ C}$ distributed uniformly. What is the magnitude of electric field at a point $15 \text{ cm}$ from the centre of the sphere? (2020) $(\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2)$
Electric field $E = \frac{kq}{r^2}$. Substituting the given values, $E = \frac{9 \times 10^9 \times 3.2 \times 10^{-7}}{(0.15)^2}$. Solving this yields $E = 1.28 \times 10^5 \text{ N/C}$.
Question 10:
easyA hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre (2019)
Inside a hollow metal sphere, the electric field is zero ($E = 0$ for $r < R$). Outside the sphere, it behaves as a point charge concentrated at the center, so $E \propto \frac{1}{r^2}$, which decreases as r increases for $r > R$.