Question

A 2.98 nF parallel-plate capacitor is charged to an initial potential difference of 49 V and then isolated. The dielectric material between the plates has a dielectric constant of 3.1. What is the potential difference of the capacitor after the dielectric is withdrawn? Answer in units of V.

Answer 1

Answer:151.9 V

Step-by-step explanation:

Given

Capacitance (C)=2.98 nF

Potential difference=49 V

dielectric strength k=3.1

Charge remains same after the removal of dielectric

thus
`Q_1=Q_2`

`Q_1=2.98* 49`

Now dielectric is removed so capacitance decreases to
`(C)/(k)`

`Q_2=(2.98)/(3.1)* V`

`V=49* 3.1=151.9 V`

Answer 2

Potential difference will be 151.9 volt

Step-by-step explanation:

We have given capacitance of the capacitor
`C=2.98nF=2.98* 10^(-9)F`

Voltage V = 49 Volt

Dielectric constant K = 3.1

We have to find the potential difference

We know that when a dielectric medium is introduced then p[otential difference is increases by k times

As the dielectric constant k = 3.1

So potential difference will be = 3.1×49 = 151.9 volt