Question

A $6 \times {10^{ - 4}}$ F parallel plate air capacitor is connected to a 500 V battery. When air is replaced by another dielectric material, $7.5 \times {10^{ - 4}}$ C charge flows into the capacitor. The value of the dielectric constant of the material is:
A) $1.5$
B) $2.0$
C) $1.0025$
D) $3.5$

Answer 1

After replacing air inside the capacitor by dielectric constant in the circuit, the voltage across the capacitor is not changing. So we can use this observation to find the dielectric constant of the material which is replaced by air in the circuit.

Complete step by step answer:
Capacitance of initial parallel plate capacitor containing air is: $C = 6 \times {10^{ - 4}}$F.
Voltage across capacitor $V = 500$ Volts.
Hence we know that the initial charge on the capacitor will be $Q = CV$.
Putting value of C and V in above equation, we get,
$Q = 7.5 \times {10^{ - 4}} \times 500$
On solving this, we get,
$Q = .375 \times {10^{ - 4}}$C
Extra charge flown $q = 7.5 \times {10^{ - 4}}$ .
Total charge on capacitor when air is replaced by another dielectric material will be Q+q,
Where $Q + q = 3007.5 \times {10^{ - 4}}C$ ,
Now, the capacitance of the capacitor ${C_f}$ will be $\dfrac{{TotalQ}}{{voltage}} = \dfrac{{Q + q}}{V}$
On solving this, we get,
${C_f} = \dfrac{{3007.5 \times {{10}^{ - 4}}}}{{500}}$ ,
${C_f} = 6.015 \times {10^{ - 4}}F$ ,
Now, let the dielectric constant of material be K, so final capacitance after replacing air by material with dielectric constant of K will be KC, so we get,
${C_f} = KC$,
On solving this we get,
$K = \dfrac{{{C_f}}}{C} = \dfrac{{6.015 \times {{10}^{ - 4}}F}}{{6 \times {{10}^{ - 4}}F}}$,
On solving this we get,
$K = 1.0025$
So option (C) is correct.

Note: After replacing air by dielectric constant of K in the capacitor, the voltage across the capacitor doesn’t change and after this, the effective capacitance of the capacitor of capacitance C become KC which shows that if $K > 1$ then there will be increase in charge in capacitor otherwise charge will decrease.