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Question
Read the following paragraph and answer the questions.
| A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain the desired value of capacitance. |
- Find the equivalent capacitance between points A and B in the given diagram.
- A dielectric slab is inserted between the plates of the parallel plate capacitor. The electric field between the plates decreases. Explain.
- A capacitor A of capacitance C, having charge Q is connected across another uncharged capacitor B of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.
OR
Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in the figure. Find an expression for the capacitance of the system.
Solution
- The given circuit can be redrawn as it is a balanced Wheatstone bridge.
i.e., `C_1/C_2 = C_3/C_4 = C/C = 1/1`
So, Cs can be neglected.
So, the equivalent capacitance between A and B,
`C_{eq} = ((1 xx 1)/(1 + 1) + (1 xx 1)/(1 + 1) + 1)C = 2C` - Total charge of the capacitor remains conserved on the introduction of a dielectric slab. Also, the capacitance of the capacitor increases to K time of its original value.
∴ CV = C'V = KC.V (since Q = Q')
⇒ `V = V/K`
Now electric field,
`E = V/d = ((V"/"K))/d`
= `(V/d)(1/K) = E/K`
So, on introducing a dielectric medium.
The electric field reduces by `(1/K)` times its original value. - (a) The initial charge on capacitor Q = CV.
This capacitor gets connected to another parallel. So total capacitance will be
C' = C + 2C = 3C
The potential difference across this combination will be VC·
`V_C = Q/C^' = Q/(3C)`
As both the capacitors are connected in parallel, therefore, the voltage of both will be the same.
(b) Charges on both capacitors,
`Q_A = CV_C = C. Q/(3C) = Q/3`
`Q_B = 2CV_C = 2C. Q/(3C) = 2/3Q`
∴ Charge lost by capacitor A = `1/3Q`
OR
The effective capacitance in series,
C = `(Kepsi_0A)/d`
C = `(C_1C_2)/(C_1 + C_2)`
`C_1 = (2Kepsi_0A)/(d"/"3), C_2 = (Kepsi_0A)/(2d"/"3)`
C = `(C_1C_2)/(C_1 + C_2)`
= `(((2Kepsi_0A xx 3)/d) ((3Kepsi_0A)/(2d)))/((Kepsi_0A)/d(6 + 3/2))`
= `(9Kepsi_0A xx 2)/(15d)`
= `(6Kepsi_0A)/(5d)`
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