Two metallic spheres of radii $1 \text{ cm}$ and $2 \text{ cm}$ are given charges $10^{-2} \text{ C}$ and $5 \times 10^{-2} \text{ C}$ respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is (1995)
Total charge $Q = 10^{-2} + 5 \times 10^{-2} = 6 \times 10^{-2} \text{ C}$. Capacitances are proportional to radii, so $C_1 : C_2 = 1 : 2$. Charge on the smaller sphere is $q_1 = Q \times \frac{C_1}{C_1 + C_2} = 6 \times 10^{-2} \times \frac{1}{3} = 2 \times 10^{-2} \text{ C}$.
A hollow metallic sphere of radius $10 \text{ cm}$ is charged such that potential of its surface is $80 \text{ V}$. The potential at the centre of the sphere would be (1994)
The electric potential inside a hollow conducting sphere is constant everywhere and is equal to the potential on its surface. Therefore, the potential at the centre is also $80 \text{ V}$.
The electrostatic force between the metal plates of an isolated parallel plate capacitor $C$ having a charge $Q$ and area $A$, is (2018)
The electrostatic force between the plates of a parallel plate capacitor is given by $F = \frac{Q^2}{2A\varepsilon_0}$. This formula shows that the force depends on charge and area but is completely independent of the distance between the plates.
A parallel plate air capacitor has capacity ‘$C$’, distance of separation between plates is ‘$d$’ and potential difference ‘$V$’ is applied between the plates force of attraction between the plates of the parallel plate air capacitor is: (2015 Pre)
The force between the plates is $F = \frac{Q^2}{2A\varepsilon_0}$. Since $C = \frac{A\varepsilon_0}{d}$ and $Q = CV$, we can write $F = \frac{(CV)^2}{2(Cd)} = \frac{C^2V^2}{2Cd} = \frac{CV^2}{2d}$.
A parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates: (2006)
When the battery is disconnected, the charge $Q$ on the plates remains constant. Capacitance $C = \frac{A\varepsilon_0}{d}$. As distance $d$ increases, $C$ decreases. Since $V = \frac{Q}{C}$, a decrease in capacitance results in an increase in potential difference.
A capacitor of capacity $C_1$ charged upto $V$ volt and then connected to an uncharged capacitor of capacity $C_2$. Then final P.D. across each will be: (2002)
The common potential $V'$ after connection is the total charge divided by the total capacitance. Initial total charge is $q = C_1V + 0$. Total capacitance is $C_1 + C_2$. Therefore, $V' = \frac{C_1V}{C_1 + C_2}$.
In polar molecules, the centers of positive and negative charges do not coincide due to asymmetric charge distribution. This inherent separation of charges gives them a permanent electric dipole moment even without an external field.
The capacitance of a parallel plate capacitor with air as medium is $6 \mu\text{F}$. With the introduction of a dielectric medium, the capacitance becomes $30 \mu\text{F}$. The permittivity of the medium is : (2020) ($\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{ m}^{-2}$)
Dielectric constant $K = \frac{C}{C_0} = \frac{30}{6} = 5$. The permittivity of the medium is $\varepsilon = K\varepsilon_0 = 5 \times 8.85 \times 10^{-12} = 44.25 \times 10^{-12} = 0.4425 \times 10^{-10} \text{ C}^2\text{N}^{-1}\text{ m}^{-2}$.
A parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. A dielectric slab of dielectric constant $K$, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect? (2015)
Since the battery is disconnected before inserting the dielectric slab, the capacitor is isolated and its total charge must remain conserved. Thus, the statement 'The charge on the capacitor is not conserved' is incorrect.
Two parallel metal plates having charges $+Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will: (2010 Mains)
The electric field between the plates in a medium is $E = \frac{\sigma}{\epsilon_{0}K}$. Since kerosene has a dielectric constant $K > 1$, the electric field will decrease.