Question

Two parallel plate capacitors, X and Y have the same area of plates and same separation between them. X has air between the plates while Y contained a dielectric medium of ε_r = 4. (0 Calculate capacitance of each capacitor if equivalent capacitance of combination is 4 μF.
(ii) Calculate the potential difference between the plates of X and Y.
(iii) What is the ratio of electrostatic energy stored in X and Y?

Solution

Given, two parallel plate capacitors X and Y having same area of plates and same seperation between them.
There is vacuum as dielectric medium in between the plates of X and, dielectric medium of dielectric constant 4 is in between the plates of Y.

i) Capacitance of X, C_X =

\frac{ε_{o} A}{d}

Capacitance of Y, C_Y= 4

\frac{ε_{o} A}{d}

the ratio of capacitances X and Y is given by

\frac{C_{Y}}{C_{X}} = 4 \Rightarrow C_{Y} = 4 C_{X}

Since, X and Y are in series combination

C_eq=

\frac{C_{X} C_{Y}}{C_{X} + C_{Y}} = 4 \Rightarrow C_{X} = 5 μ F

and, C_Y= 4(5) =20 μF.

ii) Using the formula V=Q/C
Since, V has an inverse dependance on C we have,

\frac{V_{X}}{V_{Y}} = \frac{C_{Y}}{C_{X}} = 4 \Rightarrow V_{X} = 4 V_{y}

also, given V_x + V_Y= 12

\Rightarrow 4 V_{Y} + V_{Y} = 12 \Rightarrow 5 V_{Y} = 12 \Rightarrow V_{Y =} \frac{12}{5} = 2.4 V and, V_{x} = 4 (2.4) = 9.6 V

iii) Electrostatic energy stored in the capacitor is given by U=

\frac{Q^{2}}{2 C}

.

Therefore, the ratio of the energy of capacitor X to the ratio of energy of capacitor Y is,

\frac{\frac{Q^{2}}{2 C_{X}}}{\frac{Q^{2}}{2 C_{y}}} = \frac{C_{Y}}{C_{X}} = 4 \Rightarrow \frac{U_{X}}{U_{Y}} = 4 : 1

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