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Calculate the amount of charge on capacitor of \( 4\,\mu\mathrm{F} \). The internal resistance of battery is \( 1\,\Omega \):

A capacitor of \( 50 \, \mu \text{F} \) is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is ________ \( \mu \text{C} \).

Three capacitors \( C_1 = 2\,\mu \text{F} \), \( C_2 = 6\,\mu \text{F} \) and \( C_3 = 12\,\mu \text{F} \) are connected as shown in figure. Find the ratio of the charge on capacitors \( C_1 \), \( C_2 \) and \( C_3 \) respectively:

A parallel plate capacitor with plate area '\( A \)' and distance of separation '\( d \)' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as:
\( \displaystyle \varepsilon(x) = \varepsilon_0 + kx, \text{ for } \left(0 < x \le \frac{d}{2}\right) \)
\( \varepsilon(x) = \varepsilon_0 + k(d-x), \text{ for }\left(\dfrac{d}{2} \le x \le d \right)\)

In a parallel plate capacitor set up, the plate area of capacitor is \( 2 \ \text{m}^{2} \) and the plates are separated by \( 1 \ \text{m} \). If the space between the plates are filled with a dielectric material of thickness \( 0.5 \ \text{m} \) and area \( 2 \ \text{m}^{2} \) (see figure) the capacitance of the set-up will be ________ \( \varepsilon_{0} \). (Dielectric constant of the material \( = 3.2 \)) (Round off to the Nearest Integer)

A capacitor is connected to a \( 20 \, \text{V} \) battery through a resistance of \( 10 \, \Omega \). It is found that the potential difference across the capacitor rises to \( 2 \, \text{V} \) in \( 1 \, \mu \text{s} \). The capacitance of the capacitor is_____ \( \mu \text{F} \).
Given $\ell n\left(\frac{10}{9}\right) = 0.105$

A capacitor of capacitance \( C = 1 \, \mu\text{F} \) is suddenly connected to a battery of \( 100 \, \text{volt} \) through a resistance \( R = 100 \, \Omega \). The time taken for the capacitor to be charged to get \( 50 \, \text{V} \) is: [Take \( \ln 2 = 0.69 \)]

Two capacitors of capacities \( 2C \) and \( C \) are joined in parallel and charged up to potential \( V \). The battery is removed and the capacitor capacity \( C \) is filled completely with a medium of dielectric constant \( K \). Potential difference across the capacitors will now be:

If \( q_f \) is the free charge on the capacitor plates and \( q_b \) is the bound charge on the dielectric slab of dielectric constant \( k \) placed between the capacitor plates, then bound charge \( q_b \) can be expressed as:

A parallel plate capacitor has plate area \( 100 \, \text{m}^{2} \) and plate separation of \( 10 \, \text{m} \). The space between the plates is filled up to a thickness \( 5 \, \text{m} \) with a material of dielectric constant of \( 10 \). The resultant capacitance of the system is \( x \, \text{pF}\).
The value of \( \varepsilon_{0} = 8.85 \times 10^{-12} \, \text{F·m}^{-1} \)
The value of \( x \) to the nearest integer is __________ .

A radioactive sample has an average life of \( 30 \, \text{ms} \) and is decaying. A capacitor of capacitance \( 200 \, \mu \text{F} \) is first charged and later connected with resistor \( 'R' \). If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of \( 'R' \) should be ________ \( \Omega \).

A parallel-plate capacitor with plate area \( A \) has separation \( d \) between the plates. Two dielectric slabs of dielectric constant \( K_1 \) and \( K_2 \) of same area \( \dfrac{A}{2} \) and thickness \( \dfrac{d}{2} \) are inserted in the space between the plates. The capacitance of the capacitor will be given by:

Four identical rectangular plates with length, \( l = 2 \, \text{cm} \) and breadth, \( b = \dfrac{3}{2} \, \text{cm} \) are arranged as shown in figure. The equivalent capacitance between A and C is \( \dfrac{x \, \varepsilon_{0}}{d} \). The value of \( x \) is ______. (Round off to the Nearest Integer)

A parallel plate capacitor whose capacitance \( C \) is \( 14 \, \text{pF} \) is charged by a battery to a potential difference \( V = 12 \, \text{V} \) between its plates. The charging battery is now disconnected and a porcelain plate with \( k = 7 \) is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of ______ \( \text{pJ} \). (Assume no friction)

A \( 2\,\mu F \) capacitor \( C_1 \) is first charged to a potential difference of \( 10\,V \) using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor \( C_2 \) of \( 8\,\mu F \). The charge in \( C_2 \) on equilibrium condition is ________ \( \mu C \). (Round off to the Nearest Integer)

The material filled between the plates of a parallel plate capacitor has resistivity \( 200 \, \Omega \text{m} \). The value of capacitance of the capacitor is \( 2 \, \text{pF} \). If a potential difference of \( 40 \, \text{V} \) is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is: (given the value of relative permittivity of material is \( 50 \)).

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the solid capacitor will be: (Given area of plate \( = A \))

The circuit shown in the figure consists of a charged capacitor of capacity \( 3 \, \mu\text{F} \) and a charge of \( 30 \, \mu\text{C} \). At time \( t = 0 \), when the key is closed, the value of current flowing through the \( 5 \, \text{M}\Omega \) resistor is \(' x' \, \mu\text{A} \).
The value of \( 'x' \) to the nearest integer is __________ .

A parallel plate capacitor of capacitance \( 200 \, \mu \text{F} \) is connected to a battery of \( 200 \, \text{V} \). A dielectric slab of dielectric constant \( 2 \) is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be _______ \( \text{J} \).

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JEE Main 2021 27th July Morning Shift

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the solid capacitor will be: (Given area of plate \( = A \))