CK-12-Physics - Intermediate

http://www.ck12.org Chapter 16. Electric Potential

sinceβ< 1 ,β^1 >1.

Define^1 β=k, and therefore

C=kC 0 , and recall thatC 0 =e 0 Ad, and finally,

C=e 0 kAd.

Reconnecting the Voltage Source to the Capacitor

An increase in capacitance means the capacitor can acquire more charge without an electrical breakdown, since there

is less electrostatic repulsion. The induced charges on the insulator have reduced the net electric field, and we now

can have a greaterQfor the sameV.

Consider the following example:

A capacitor is charged by a 12-Volt battery. The battery is now removed and a dielectric is inserted between the

plates of the capacitor. We now measure the new potential difference between the plates and it found to be 4 V.

The 12-V battery is now reconnected to the capacitor and the potential difference between the plates of the capacitor

quickly increases to 12 V. The plates have now acquired more charge.

The same amount of charge on an air-gap capacitor could result in an electrical breakdown. There would be too

much electrostatic repulsion between like charges, and the plates would discharge.

One of the uses of capacitors is to block the flow of charge in order to protect delicate circuitry. If a sudden increase

in current were to occur in an electrical circuit, a capacitor with a dielectric would have a much better chance of

handling the increase in voltage without an electrical breakdown.

Table16.1 gives dielectric constant values for some common dielectrics. To learn more about dielectrics, follow the

link below.

http://www.youtube.com/watch?v=e0n6xLdwaT0

TABLE16.1:

Material Dielectric Constant(k)Att= 20 ◦C

Vacuum 1.0000

Air 1.0006

Rubber 2.8

Vinyl 2.8 - 4.5

Paper 3 - 7

Glass 4 - 7

Water 80

Titanium Oxide 100

Check Your Understanding

A capacitor has a layer of rubber inserted between its conducting plates while connected to a battery. By what factor

will the charge on the capacitor increase? The voltage on the capacitor remains constant.

Answer:We can answer this by just looking at the table above. The capacitance of rubber is 2.8 times greater than

air, so we should expect the charge on the capacitor to be 2.8 times greater, as well. Using the equationQ 0 =C 0 V,

whereC 0 is the capacitance of air andQ 0 is the original amount of charge on the capacitor,

Q 0 =C 0 V→C=KC 0 = 2. 8 C 0 →Q= 2. 8 (C 0 V)→ 2. 8 Q 0 →Q= 2. 8 Q 0