# 16.3 Capacitors  (Page 2/3)

 Page 2 / 3

## Physical properties of the capacitor and capacitance

The capacitance of a capacitor is proportional to the surface area of the conducting plate and inversely proportional to the distancebetween the plates. It also depends on the dielectric between the plates. We say that it is proportional to the permittivity of the dielectric . The dielectric is the non-conducting substance that separates the plates. As mentioned before, dielectrics can be air, paper, mica, perspexor glass.

The capacitance of a parallel-plate capacitor is given by:

$C={ϵ}_{0}\frac{A}{d}$

where ${ϵ}_{0}$ is, in this case, the permittivity of air, $A$ is the area of the plates and $d$ is the distance between the plates.

What is the capacitance of a capacitor in which the dielectric is air, the area of the plates is $0,001{\mathrm{m}}^{2}$ and the distance between the plates is $0,02\mathrm{m}$ ?

1. We need to determine the capacitance of the capacitor.

2. We can use the formula:

$C={ϵ}_{0}\frac{A}{d}$
3. We are given the area of the plates, the distance between the plates and that the dielectric is air.

4. $\begin{array}{ccc}\hfill C& =& {ϵ}_{0}\frac{A}{d}\hfill \\ & =& \frac{\left(8,9×10-12\right)\left(0,001\right)}{0,02}\hfill \\ & =& 4,45×{10}^{-13}\mathrm{F}\hfill \end{array}$

## Electric field in a capacitor

The electric field strength between the plates of a capacitor can be calculated using the formula:

$E=\frac{V}{d}$

where $E$ is the electric field in $\mathrm{J}.{\mathrm{C}}^{-1}$ , $V$ is the potential difference in volts ( $\mathrm{V}$ ) and $d$ is the distance between the plates in metres ( $\mathrm{m}$ ).

What is the strength of the electric field in a capacitor which has a potential difference of $300\mathrm{V}$ between its parallel plates that are $0,02\mathrm{m}$ apart?

1. We need to determine the electric field between the plates of the capacitor.

2. We can use the formula:

$E=\frac{V}{d}$

3. We are given the potential difference and the distance between the plates.

4. $\begin{array}{ccc}\hfill E& =& \frac{V}{d}\hfill \\ & =& \frac{300}{0,02}\hfill \\ & =& 1,50×{10}^{4}\mathrm{J}.{\mathrm{C}}^{-1}\hfill \end{array}$

## Capacitance and the parallel plate capacitor

1. Determine the capacitance of a capacitor which stores $9×{10}^{-9}\mathrm{C}$ when a potential difference of 12 V is applied to it.
2. What charge will be stored on a $5\mu \mathrm{F}$ capacitor if a potential difference of $6\mathrm{V}$ is maintained between its plates?
3. What is the capacitance of a capacitor that uses air as its dielectric if it has an area of $0,004{\mathrm{m}}^{2}$ and a distance of $0,03\mathrm{m}$ between its plates?
4. What is the strength of the electric field between the plates of a charged capacitor if the plates are $2\mathrm{mm}$ apart and have a potential difference of $200\mathrm{V}$ across them?

run demo

## A capacitor in a circuit

When a capacitor is connected in a DC circuit, current will flow until the capacitor is fully charged. After that, no furthercurrent will flow. If the charged capacitor is connected to another circuit with no source of emf in it, the capacitor will discharge through the circuit, creating a potential difference fora short time. This is useful, for example, in a camera flash.

Initially, the electrodes have no net charge. A voltage source is applied to charge a capacitor. The voltage source creates an electricfield, causing the electrons to move. The charges move around the circuit stopping at the left electrode. Here they are unable totravel across the dielectric, since electrons cannot travel through an insulator. The charge begins to accumulate, and anelectric field forms pointing from the left electrode to the right electrode. This is the opposite direction of the electric fieldcreated by the voltage source. When this electric field is equal to the electric field created by the voltage source, the electronsstop moving. The capacitor is then fully charged, with a positive charge on the left electrode and a negative charge on the rightelectrode.

during a snooker competition ,a 200g ball A m moving with velocity va collide head on with a identical ball B that was at rest.A after the collision ball A remains at rest wile ball B moves on with a velocity of 4m/s? With what speed was ball a moving before the collision
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