CK-12-Physics - Intermediate

16.4. Dielectrics http://www.ck12.org

FIGURE 16.14

The capacitor with the dielectric inserted showing the polarization of

charges within the dielectric.

FIGURE 16.15

The change of the electric field within the capacitor with the dielectric

inserted.

A Mathematical Explanation of

The chargeQ 0 on each side of the capacitor remains constant since the capacitor is disconnected from the battery.

Thus, ifC 0 is the original capacitance andV 0 is the original voltage of the capacitor, then

Q 0 =C 0 V 0 (Equation A)

After the dielectric has been inserted into the capacitor, the new voltage is less than the originalV 0 →Vvoltage,

whereV<V 0. But the product of capacitance and voltage must equal the same chargeQ 0. Thus, if the voltage

decreases, the capacitance increases, sinceQ 0 remains constant.

Q 0 =CV(Equation B)

Setting Equation A equal to Equation B and solving forC:

C 0 V 0 =CV→C=C 0 VV^0 (Equation C)

We defineV=βV 0 , whereβ<1, and substituteβV 0 forVin Equation C.

C=C (^0) βVV^00 =C (^01) β,