Force between Plates of a Capacitor

The charges on two parallel copper plates separated by a small distance are $+Q$ and $-Q$. A test charge q experiences a force $F$ when placed between these plates. Now if one of the plates is removed to infinity, then the force on the test charge will become-

One plate of a capacitor is connected to a spring as shown in figure. Area of both the plates is $A$. In steady state separation between the plates is $0.8d$ (spring was unstretched and the distance between the plates was $d$ when the capacitor was uncharged). The force constant of the spring is approximately:

The force between the plates of a parallel plate capacitor of capacitance $C$ and distance of separation of the plates $d$ with a potential difference $V$ between the plates, is-

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

If an electron enters into a space between the plates of a parallel plate capacitor at an angle $\mathrm{\alpha }$ with the plates and leaves at an angle $\mathrm{\beta }$ to the plates. The ratio of its kinetic energy while entering the capacitor to that while leaving will be

Two large non conducting plates having surface charge densities $+\sigma and-\sigma$ respectively, are fixed d distance apart. A small test charge q of mass m is attached to two non conducting springs each of spring constant k as shown in the figure. The sum of lengths of both springs in undeformed state is $d$. The charge $q$ is released from rest with both the springs nondeformed. Then charge $q$ will (neglect gravity)

A parallel plate capacitor has an electric field of ${10}^5\mathrm{ }\mathrm{V} {\mathrm{m}}^{-1}$ between the plates. If the charge on the capacitor plate is $1 \mathrm{\mu }\mathrm{C}$, the force on each capacitor plate is:

In the given figure a capacitor of plate area $\mathrm{A}$ is charged up to charge $\mathrm{q}$. The mass of each plate is $m_2$. The lower plate is rigidly fixed. Find the value of $m_1$ so that the system remains in equilibrium -

Two charged capacitor have their outer plates fixed and inner plates connected by a spring of force constant $\mathrm{k}$ . The charge on each capacitor is $\mathrm{q}$ . Find the extension in the spring at equilibrium-

In the given system a capacitor of plate area $A$ is charged up to charge $q$ . The mass of each plate is $m_2$ . The lower plate is rigidly fixed. Find the value of $m_1$ so that the system is in equilibrium -
 

A parallel plate capacitor having charge $Q$on it and plate area $A$, has $d$ separation between the plates. If one of the plates is pulled to make the final separation $2d$, then work done in this process is:

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2-V_1=20$$V$. (,$ie$, plate $2$ is at a higher potential). The plates are separated by $d=0.1$$m$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $1$. What is its speed when it hits plate 2? $e$(=1.6$\times {10}^{-19}\mathrm{C},m_0=9.11\times {10}^{-31}\mathrm{k}\mathrm{g})$

A capacitor is connected to a battery. When separation between them is halved, the force of attraction between the plates

In the given figure capacitor of plate area A is charged upto charge q. The ratio of elongation (neglect force of gravity) in spring C and D at equilibrium position is

One plate of a capacitor is connected to a spring as shown in the igure. Area of both the plates is $A$. In steady-state, the separation between the plates is $0.8 d$ (spring was unstretched and the distance between the plates was $d$ when the capacitor was uncharged). The force constant of the spring is