A capacitor connected to a $10$ V battery collects a charge $40 \mu C$ with air as dielectric and $100 \mu C$ with oil as dielectric. The dielectric constant of oil is
Answer
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Hint: The capacitor contains two parallel plates that are not connected but are electrically segregated either by air or by some dielectric material with a dielectric constant K. The capacitance is the capacitor's ability to collect the electric charge, and it is determined by dividing the electric charge by voltage. Its unit is Farad.
Formula used:
$C = \dfrac{Q}{V}$
Where C is capacitance, V is voltage applied and Q is the charge collected.
Capacitance when dielectric material is air-
$C = \dfrac{\epsilon_{0} A}{d}$
Where, C is capacitance, $\epsilon_{0}$ is permittivity, A is the parallel plate area and d is the distance between the two plates
Capacitance when dielectric material is any other material-
$C = \dfrac{K \epsilon_{0} A}{d}$
K is the dielectric constant.
Complete step-by-step answer:
Case 1- when dielectric material is air.
Put Q = $40 \mu C$ = $40 \times 10^{-6} $C; V=V in the formula given below: -
$C_{1} = \dfrac{Q}{V} = \dfrac{40 \times 10^{-6} }{10} = 4 \times 10^{-6}$ F
Capacitance of parallel plates when dielectric material is air.
$C_{1} = \dfrac{K \epsilon_{0} A}{d}$
Case 2- when dielectric material is oil.
Put Q = $100 \mu C$ = $100 \times 10^{-6} $C; V=V in the formula given below: -
$C_{2} = \dfrac{Q}{V} = \dfrac{100 \times 10^{-6} }{10} = 10 \times 10^{-6}$ F
Capacitance of parallel plates when dielectric material is oil.
$C_{2} = \dfrac{K \epsilon_{0} A}{d}$
On diving $C_{2}$ by $C_{1}$, we get-
$K=\dfrac{ C_{2}}{ C_{1}}$
$K=\dfrac{10 \times 10^{-6} }{ 4 \times 10^{-6}} = 2.5 $
Dielectric constant of oil is $2.5$.
Additional Information: The factor by which the dielectric substance, or insulator, raises the capacitor's capacitance compared to air is called the Dielectric Constant. A dielectric material with a high K is more suited to insulators than a dielectric material with a low K. The dielectric constant is a dimensionless term.
Note: A capacitor has a charge Q, which means the positively charged conductor has a charge +Q, and the negatively charged conductor has a charge -Q, but the total charge on the capacitor is zero. The conductor with positively charged is at a higher potential than the conductor with negatively charged.
Formula used:
$C = \dfrac{Q}{V}$
Where C is capacitance, V is voltage applied and Q is the charge collected.
Capacitance when dielectric material is air-
$C = \dfrac{\epsilon_{0} A}{d}$
Where, C is capacitance, $\epsilon_{0}$ is permittivity, A is the parallel plate area and d is the distance between the two plates
Capacitance when dielectric material is any other material-
$C = \dfrac{K \epsilon_{0} A}{d}$
K is the dielectric constant.
Complete step-by-step answer:
Case 1- when dielectric material is air.
Put Q = $40 \mu C$ = $40 \times 10^{-6} $C; V=V in the formula given below: -
$C_{1} = \dfrac{Q}{V} = \dfrac{40 \times 10^{-6} }{10} = 4 \times 10^{-6}$ F
Capacitance of parallel plates when dielectric material is air.
$C_{1} = \dfrac{K \epsilon_{0} A}{d}$
Case 2- when dielectric material is oil.
Put Q = $100 \mu C$ = $100 \times 10^{-6} $C; V=V in the formula given below: -
$C_{2} = \dfrac{Q}{V} = \dfrac{100 \times 10^{-6} }{10} = 10 \times 10^{-6}$ F
Capacitance of parallel plates when dielectric material is oil.
$C_{2} = \dfrac{K \epsilon_{0} A}{d}$
On diving $C_{2}$ by $C_{1}$, we get-
$K=\dfrac{ C_{2}}{ C_{1}}$
$K=\dfrac{10 \times 10^{-6} }{ 4 \times 10^{-6}} = 2.5 $
Dielectric constant of oil is $2.5$.
Additional Information: The factor by which the dielectric substance, or insulator, raises the capacitor's capacitance compared to air is called the Dielectric Constant. A dielectric material with a high K is more suited to insulators than a dielectric material with a low K. The dielectric constant is a dimensionless term.
Note: A capacitor has a charge Q, which means the positively charged conductor has a charge +Q, and the negatively charged conductor has a charge -Q, but the total charge on the capacitor is zero. The conductor with positively charged is at a higher potential than the conductor with negatively charged.
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