Question

A paper filled capacitor is charged to a potential difference of 2.1 V and then disconnected from the charging circuit. The dielectric constant of the paper is 3.7. Keeping the plates insulated, the paper filling is withdrawn, allowing air to fill the capacitor instead. Find the resulting potential difference off the capacitor.
(volts)
While continuing to keep the capacitor's plates insulated, an unknown substance is inserted between them. The plates then attain a potential difference that is 0.59 times the original potential difference (when paper filled the capacitor). What is the substances dielectric constant?

Answer 1

Answer:

Part a)

$V' = 7.77 Volts$

Part b)

$k' = 6.27$

Explanation:

As we know that capacitor plate is connected across 2.1 V and after charging it is disconnected from the battery

So here we can say that charge on the plates will remain conserved

So we will have

$Q = kC(2.1)$

now dielectric is removed between the plates of capacitor

so new potential difference between the plates

$V' = \frac{Q}{C'}$

$V' = \frac{kC(2.1)}{C}$

$V' = 3.7 \times 2.1$

$V' = 7.77 Volts$

Part b)

Now the capacitor plates are again isolated and unknown dielectric is inserted between the plates

So again charge is same so potential difference is given as

$V" = \frac{Q}{k'C}$

$0.59 V = \frac{kCV}{k'C}$

$0.59 = \frac{3.7}{k'}$

$k' = 6.27$