A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as $\varepsilon=\alpha U$ where $a=2 \mathrm{~V}^{-1}$. A similar capacitor with no dielectric is charged to $U_0=78 \mathrm{~V}$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.
Assuming the required final voltage be $U$. If $C$ is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is given by $Q_1=C U$
Since, the capacitor with the dielectric has a capacitance $\varepsilon C$. Hence, the charge on the capacitor is given by
$$Q_2=\varepsilon C U=(\alpha U) C U=\alpha C U$$
The initial charge on the capacitor is given by
$$Q_0=C U_0$$
From the conservation of charges,
$$\begin{aligned} Q_0 & =Q_1+Q_2 \\ \text{Or}\quad C U_0 & =C U+\alpha C U^2 \end{aligned}$$
$$\begin{aligned} &\begin{aligned} & \Rightarrow \quad \alpha U^2+U-U_0=0 \\ & \therefore \quad U=\frac{-1 \pm \sqrt{1+4 \alpha U_0}}{2 \alpha} \end{aligned}\\ &\text { On solving for } U_0=78 \mathrm{~V} \text { and } a=2 / \mathrm{V} \text {, we get }\\ &U=6 \mathrm{~V} \end{aligned}$$
A capacitor is made of two circular plates of radius $R$ each, separated by a distance $d \ll R$. The capacitor is connected to a constant voltage. A thin conducting disc of radius $r \ll R$ and thickness $t \ll r$ is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is $m$.
Assuming initially the disc is in touch with the bottom plate, so the entire plate is a equipotential.
The electric field on the disc, when potential difference $V$ is applied across it, given by
$$E=\frac{V}{d}$$
Let charge $q^{\prime}$ is transferred to the disc during the process,
Therefore by Gauss' theorem,
$$\begin{aligned} \therefore\quad q^{\prime} & =-\varepsilon_0 \frac{V}{d} \pi r^2 \\ \text{Since, Gauss theorem states that}\quad \phi & =\frac{q}{\varepsilon_0} \text { or } q=\frac{\varepsilon_0}{\phi} \\ & =\varepsilon E A=\frac{\varepsilon_0 V}{d} A \end{aligned}$$
The force acting on the disc is
$$-\frac{V}{d} \times q^{\prime}=\varepsilon_0 \frac{V^2}{d^2} \pi r^2$$
If the disc is to be lifted, then
$$\varepsilon_0 \frac{V^2}{d^2} \pi r^2=m g \Rightarrow V=\sqrt{\frac{m g d^2}{\pi \varepsilon_0 r^2}}$$
This is the required expression.
(a) In a quark model of elementary particles, a neutron is made of one up quarks [charge $(2 / 3) e$ ] and two down quarks [charges - (1/3)e]. Assume that they have a triangle configuration with side length of the order of $10^{-15} \mathrm{~m}$. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV .
(b) Repeat above exercise for a proton which is made of two up and one down quark.
This system is made up of three charges. The potential energy of the system is equal to the algebraic sum of PE of each pair. So,
$$\begin{aligned} U & =\frac{1}{4 \pi \varepsilon_0}\left\{\frac{q_d q_d}{r}-\frac{q_u q_d}{r}-\frac{q_u q_d}{r}\right\} \\ & =\frac{9 \times 10^9}{10^{-15}}\left(1.6 \times 10^{-19}\right)^2\left[\left\{(1 / 3)^2-(2 / 3)(1 / 3)-(2 / 3)(1 / 3)\right\}\right] \\ & =2.304 \times 0^{-13}\left\{\frac{1}{9}-\frac{4}{9}\right\}=-7.68 \times 10^{-14} \mathrm{~J} \\ & =4.8 \times 10^5 \mathrm{eV}=0.48 \mathrm{meV}=5.11 \times 10^{-4}\left(m_n c^2\right) \end{aligned}$$
Two metal spheres, one of radius $R$ and the other of radius $2 R$, both have same surface charge density $\sigma$. They are brought in contact and separated. What will be new surface charge densities on them?
The charges on two metal spheres, before coming in contact, are given by
$$\begin{aligned} Q & =\sigma .4 \pi R^2 \\ Q_2 & =\sigma .4 \pi\left(2 R^2\right) \\ & =4\left(\sigma .4 \pi R^2\right)=4 Q_1 \end{aligned}$$
Let the charges on two metal spheres, after coming in contact becomes $Q_1^{\prime}$ and $Q_2^{\prime}$.
Now applying law of conservation of charges is given by
$$\begin{aligned} Q_1^{\prime}+Q_2^{\prime} & =Q_1+Q_2=5 Q_1 \\ & =5\left(\sigma \cdot 4 \pi R^2\right) \end{aligned}$$
After coming in contact, they acquire equal potentials. Therefore, we have
$$\frac{1}{4 \pi \varepsilon_0} \frac{Q_1^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{Q_2^{\prime}}{R}$$
On solving, we get
$$\begin{array}{ll} \therefore & Q_1^{\prime}=\frac{5}{3}\left(\sigma .4 \pi R^2\right) \text { and } Q_2^{\prime}=\frac{10}{3}\left(\sigma .4 \pi R^2\right) \\ \therefore & \sigma_1=5 / 3 \sigma \text { and } \\ \therefore & \sigma_2=\frac{5}{6} \sigma \end{array}$$
In the circuit shown in figure, initially $K_1$ is closed and $K_2$ is open. What are the charges on each capacitors? Then $K_1$ was opened and $K_2$ was closed (order is important), what will be the charge on each capacitor now? $[C=1 \propto \mathrm{~F}]$
In the circuit, when initially $K_1$ is closed and $K_2$ is open, the capacitors $C_1$ and $C_2$ acquires potential difference $V_1$ and $V_2$ respectively. So, we have
$$\begin{aligned} &\text { and }\\ &\begin{aligned} & V_1+V_2=E \\ & V_1+V_2=9 V \end{aligned} \end{aligned}$$
Also, in series combination ,
$$\begin{aligned} V & \propto 1 / C \\ V_1: V_2 & =1 / 6: 1 / 3 \end{aligned}$$
On solving
$$\begin{array}{ll} \Rightarrow & V_1=3 V \text { and } V_2=6 \mathrm{~V} \\ \therefore & Q_1=C_1 V_1=6 \mathrm{C} \times 3=18 \propto \mathrm{C} \\ & Q_2=9 \propto \mathrm{C} \text { and } Q_3=0 \end{array}$$
Then, $K_1$ was opened and $K_2$ was closed, the parallel combination of $C_2$ and $C_3$ is in series with $C_1$.
$$Q_2=Q_2^{\prime}+Q_3$$
and considering common potential of parallel combination as $V$, then we have
$$\begin{aligned} C_2 V+C_3 V & =Q_2 \\ \Rightarrow\quad V & =\frac{Q_2}{C_2+C_3}=(3 / 2) \mathrm{V} \\ \text{On solving,}\quad Q_2^{\prime} & =(9 / 2) \propto \mathrm{C} \\ \text{and}\quad Q_3 & =(9 / 2) \propto \mathrm{C} \end{aligned}$$