Question

Two parallel plate capacitors X and Y have the same area of the plates and the same separation between them is connected in series to a battery of 15 V. X has air between the plates while Y contains a dielectric of constant k = 2.

i) Calculate the capacitance of each capacitor if the equivalent capacitance of the combination is 2 μF.

ii) Calculate the potential difference between the plates of X and Y.

iii) What is the ratio of the electrostatic energy stored in X and Y?  

Answer 1

Given: V = 15 volt, k_X = 1, k_Y = 2, C_s = C_eq = 2 μF 

To Find:

1. C_X and C_Y 
2. V_X and V_Y 
3. `"E"_"X"/"E"_"Y"`

Formulae:

1. C = `(ε_0"kA")/"d"`
2. `1/"C"_"s" = 1/"C"_1 + 1/"C"_2`
3. Q = CV
4. E = `"Q"^2/(2"C")`

Calculation:

From formula (i),

C ∝ k 

∴ `"C"_"X"/"C"_"Y" = "k"_"X"/"k"_"Y" = 1/2`

∴ C_Y = 2C_X     ….(1)

∴ From formula (ii),

`1/"C"_"eq" = 1/"C"_"X" + 1/"C"_"Y"`

`1/(2mu"F") = 1/"C"_"X" + 1/(2"C"_"X")`

∴ C_x = 3 μF

From equation (1),

C_Y = 6 μF

From formula (iii), 

Q = `"C"_"eq" xx "V"`

= `2 xx 10^-6 xx 15`

= 30 μC 

∴ `"V"_"x" = "Q"/"C"_"X" = (30 xx 10^-6)/(3 xx 10^-6)` = 10 V 

∴ `"V"_"y" = "Q"/"C"_"Y" = (30 xx 10^-6)/(6 xx 10^-6)` = 5 V 

From formula (iv),

`"E"_"X"/"E"_"Y" = "Q"^2/(2"C"_"X") xx (2"C"_"Y")/"Q"^2`

= `"C"_"Y"/"C"_"X" = (6 xx 10^-6)/(3 xx 10^-6) = 2`

1. Capacitance of each capacitor are 3 μF and 6 μF.
2. The potential difference between the plates of X and Y is 10 V and 5 V respectively.
3. The ratio of electrostatic energy stored in X and Y is 2:1.