Parallel Plate Capacitor

Two circular plates each of radius '$r$' are kept parallel to each other distance '$d$' apart. The capacitance of the capacitor formed is '$C_1$ '. If the radius of each of the plates is increased to $\sqrt{3}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes '$C_2$'. The ratio $C_1 : C_2$ is

Two indentical metal plates are given charges $q_1$ and $q_2\ (q_2 < q_1)$ respectively. They are brought close together to form a parallel plate capacitor with capacitance 'C'. The potential difference 'V' between the plates is

A parallel plate capacitor was made with two rectangular plates, each with a length of $l = 3\text{ cm}$ and breadth of $b = 1\text{ cm}$. The distance between the plates is $3\ \mu\text{m}$. Out of the following, which are the ways to increase the capacitance by a factor of 10?

A. $l = 30\text{ cm},\ b = 1\text{ cm},\ d = 1\ \mu\text{m}$

B. $l = 3\text{ cm},\ b = 1\text{ cm},\ d = 30\ \mu\text{m}$

C. $l = 6\text{ cm},\ b = 5\text{ cm},\ d = 3\ \mu\text{m}$

D. $l = 1\text{ cm},\ b = 1\text{ cm},\ d = 10\ \mu\text{m}$

E. $l = 5\text{ cm},\ b = 2\text{ cm},\ d = 1\ \mu\text{m}$

Choose the correct answer from the options given below:

Two identical metal plates are given charges $q_1$ and $q_2\ (q_2 < q_1)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance ' C ', the potential difference ' $V$ ' between the plates is

An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity $10^{6}\text{ m/s}$. If the magnitude of the electric field between the plates is $9.1\text{ V/cm}$, then the vertical component of velocity of electron is (mass of electron $= 9.1 \times 10^{-31}\text{ kg}$ and charge of electron $= 1.6 \times 10^{-19}\text{ C}$)

Two circular metal plates each of radius ' $r$ ' are kept parallel to each other distance ' $d$ ' apart. The capacitance of the capacitor formed is ' $C_1$ '. If the radius of each of the plates is increased to $\sqrt{2}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes ' $C_2$ '. The ratio of the capacitances $C_1 : C_2$ is

Charge on a parallel plate capacitor of capacity C is Q, the electric field intensity between its two plates separated by a distance of $t$ is

A capacitor of capacitance $C$ is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is:

The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If C is its initial capacitance, its final capacitance is equal to

Two identical thin metal plates have charges $q_1$ and $q_2$ respectively such that $q_1 > q_2$. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is:

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

The plates of a parallel plate capacitor of capacity '$C_1$' are moved closer together until they ant half their original separation. The new capacitance '$C_2$' is

A simple pendulum of mass 'm', length 'l' and charge '+ q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be

An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle '$\alpha$' with the plates. It leaves the plates at angle '$\beta$' with kinetic energy K2. Then the ratio of kinetic energies K1: K2 will be:

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$ between them, as shown in figure. The capacitance will be close to:

A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is:

(Take $\varepsilon$0 = 8.85 $\times$ 10$-$12 $\frac{C^2}{N\text{-}m^2}$)

A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μC charge and the other plate, +4μC charge. The potential difference developed across the capacitor is:-

A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by: