List of top Physics Questions on Capacitors and Capacitance asked in MHT CET

A parallel plate capacitor has capacitance $C$. If a dielectric slab of dielectric constant $K$ and thickness equal to half the plate separation is introduced parallel to the plates, the new capacitance is:
Calculate the energy stored in a \(10\,\mu F\) capacitor charged to \(50\,V\).
Find the energy stored in a capacitor of \(10\,\mu F\) charged to a potential of \(50\,V\).
Determine the equivalent capacitance of an infinite circuit formed by repeating identical capacitors of capacitance \(C\).
Calculate the equivalent capacitance of an infinite circuit formed by repeating identical capacitors of capacitance \(C\).
Two parallel plate air capacitors of same capacity ' C ' are connected in parallel to a battery of e.m.f. ' $E$ '. Then one of the capacitors is completely filled with dielectric material of constant ' K '. The change in the effective capacity of the parallel combination is
The potential difference across the (4 \muF) capacitor in the following circuit is
A series combination of $10$ capacitors, each of value ' $C_1$ ' is charged by a source of potential difference ' $4\text{ V}$ '. When another parallel combination of $8$ capacitors, each of value ' $C_2$ ' is charged by a source of potential difference ' $V$ ', it has the same total energy stored in it as in the first combination. The value of ' $C_2$ ' is
The potential difference across the (4 \muF) capacitor in the following circuit is
Two parallel plate air capacitors of same capacity ' C ' are connected in parallel to a battery of e.m.f. ' $E$ '. Then one of the capacitors is completely filled with dielectric material of constant ' K '. The change in the effective capacity of the parallel combination is
The potential difference across the (4 \muF) capacitor in the following circuit is
Energy stored in a capacitor is given by the equation \[ E = \frac{1}{2} C V^2 \] where: - \( C \) is the capacitance, - \( V \) is the voltage, - \( E \) is the energy stored. Given the values of \( C \), \( V \), and \( E \), determine the energy stored.}
A parallel combination of two capacitors of capacities ' \(C\) ' and ' \(\frac{C}{3}\) ' respectively is connected across a battery of 12 volt. When both capacitors are fully charged, the charge and energy stored in them is \(Q_1, Q_2\) and \(E_1, E_2\) respectively. Then the ratio of \((E_1 - E_2)\) to \((Q_1 - Q_2)\) is
The ratio of potential difference that must be applied across parallel and series combination of two capacitors $C_{1}$ and $C_{2}$ with their capacitance in the ratio 1: 2 so that energy stored in these two cases becomes same is
The potential energy of charged parallel plate capacitor is $v_{0}$. If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be
The capacitance of a parallel plate capacitor is 2.5 µF. When it is half filled with a dielectric as shown in figure, its capacitance becomes 5 μF. The dielectric constant of the dielectric is
Two condensers one of capacity $\frac{C}{2}$ and other capacity $C$ are connected to a battery of voltage $V$ as shown. The work done in charging fully both the condensers is
A parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu\text{F}$. The area of capacitor is divided into two equal halves and filled with two media (as shown in figure) having dielectric constant $K_1 = 2$ and $K_2 = 4$. The capacitance of the system will be
A parallel combination of two capacitors of capacities ' \(C\) ' and ' \(\frac{C}{3}\) ' respectively is connected across a battery of 12 volt. When both capacitors are fully charged, the charge and energy stored in them is \(Q_1, Q_2\) and \(E_1, E_2\) respectively. Then the ratio of \((E_1 - E_2)\) to \((Q_1 - Q_2)\) is
Which of the following combination of 7 identical capacitors each of $2\,\mu F$ gives a capacitance of $\frac{10}{11}\,\mu F$?