Question

The potential difference between the plates of a
parallel-plate capacitor is 200 V. The area of each plate
is 100 cm² and the distance between them is 1 mm.
If the medium between them is air, then calculate the
charge taken by the capacitor. If there is a medium of
dielectric constant 2.5 between the plates, then what
will be the potential difference for the same charge?
​

Answer 1

Answer:

As a very first step, one could read the question well and hence note down the given values. Then consider the first case then recall the relations accordingly and then perform the substitutions accordingly and hence find the answer. Do the same for the second part.

Formula used:

Capacitance,

C=Aε0d

Complete step-by-step solution:

So, we have the potential difference between the plates to be 200V and the area of each plate to be 100cm2and the distance between them is 1mm.

So, we have the capacitance to be given by,

C=Aε0d=10−2×8.85×10−1210−3=8.85×10−11F

Now we have the charge taken by the capacitor for air as medium would be,

Q=CV=8.85×200×10−11=1.76×10−8C

Now, a medium of dielectric constant 2.5 is introduced between the plates and we are asked to find the potential difference for the same charge under this condition.

Now, for the same charge we have,

C1V1=C2V2

Also, the capacitance would now be,

C2=kC1

⇒V2=C1kC1×V1=V1k=2002.5

∴V2=800V

Therefore, we found the potential difference on introduction of the new medium between the plates to be 800V.

Note: So, you may have noted that we have simply multiplied the dielectric constant to the initial capacitance to get the capacitance for the second case. Now, from the expression for capacitor with a medium of dielectric constant k is given by,

C=kAε0d

In this expression, both A and d remain the same for the second part and hence the value of the capacitance.

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