Question

Ch 17 P49. An uncharged capacitor is connected to a 21.0-V battery until it is fully charged, after which it is disconnected from the battery. A slab of paraffin is then inserted between the plates. What will now be the voltage between the plates?

Answer 1

V = 9.33 V

Step-by-step explanation:

By definition, the capacitance of a capacitor, is the proportion between the charge on one of charged surfaces and the potential difference between these surfaces, as follows:

`C = (Q)/(V)`

for a parallel-plate type capacitor, it can be showed, applying Gauss'Law to the a gaussian surface with the shape of a pillbox parallel to the surface of one of the plates, half outside the surface, half inside it, that the capacitance can be expressed as follows:

`C =(\epsilon*A)/(d)`

As we can see, if a slab of paraffin, with dielectric constant of 2.25, is inserted, the capacitance will be increased in the same factor:

`Cf = 2.25 * Co`

Now, if the capacitor, once charged, is disconnected from the battery, the charge Q will remain constant.

So, if the capacitance will be increased 2.25 times, the only way to do this is that the voltage between plates be reduced in the same factor:

`Cf = 2.25 * Co = 2.25* (Q)/(Vo) = (Q)/(Vf) \\ Vf= (Vo)/(2.25) =(21.0V)/(2.25) = 9.33 V`