Question 21:
easyForce between two identical spheres charged with same charge is \(F \). If 50% charge of one sphere is transferred to the other sphere then the new force will be:
Initially, \( F = k \frac{q^2}{r^2}\). When 50% of the charge of one sphere is transferred to the other, the charges become \(0.5q\) and
\(1.5q\). The new force is \(F' = k \frac{(0.5q)(1.5q)}{r^2} = 0.75 F = \frac{3}{4}F\).
Question 22:
easyTwo point charges exert a force \(F_0\) on each other when placed in vacuum. Now the charges are increased to four times, separation between them is doubled and the system is placed is an insulating medium. Now they experience the same force. What should be the dielectric constant of the medium?
Initial force in vacuum \(F_0 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . New force in medium \(F' = \frac{1}{4\pi\epsilon_0 K} \frac{(4q_1)(4q_2)}{(2r)^2} = \frac{16}{4K} \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . Since \(F' = F_0\), we have \(frac{4}{K} = 1\), so \(K = 4\).
Question 23:
difficultTwo positive charges \(q_1\) and \(q_2\) having their sum \(Q\) are placed at \(d\) distance apart. For what values of the charges, the coulomb force between them will be maximum?
Coulomb force is proportional to the product \(q_1 q_2\). Since the sum \(q_1 + q_2 = Q\) is constant, the product is maximum when the charges are equal, i.e., \(q_1 = q_2 = Q/2\).
Question 24:
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 25:
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 26:
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 27:
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 28:
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 29:
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 30:
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 $q = -Q/4$.