Electrostatic Potential and Capacitance
- What is the effective capacitance between points X and Y?
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Equivalent circuit
Here, C1 = C2 C3 C4
Hence, no charge will flow through 20µF
C1 and C234 Hence, C' = 3 µF, C'' = 3 µF C' and C'' are in parallel.
Hence net capacitance = C' + C'' = 3 + 3
= 6 µFCorrect Option: D
Equivalent circuit
Here, C1 = C2 C3 C4
Hence, no charge will flow through 20µF
C1 and C234 Hence, C' = 3 µF, C'' = 3 µF C' and C'' are in parallel.
Hence net capacitance = C' + C'' = 3 + 3
= 6 µF
- A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes .
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When oil is placed between space of plates
C = 2Aε0 ..........(i) ∵ C = KAε0 , if K = 2 d d When oil is removed C' = Aε0 ......(ii) d
On comparing both equations, we get C ' = C/2Correct Option: D
When oil is placed between space of plates
C = 2Aε0 ..........(i) ∵ C = KAε0 , if K = 2 d d When oil is removed C' = Aε0 ......(ii) d
On comparing both equations, we get C ' = C/2
- A capacitor is charged to store an energy U. The charging battery is disconnected. An identical capacitor is now connected to the first capacitor in parallel. The energy in each of the capacitor is
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In Ist case when capacitor C connected with battery charged with the energy,
U1 = U (stored energy on capacitor).
In IInd case, after disconnection of battery similar capacitor is connected in parallel with Ist capacitor then Ceq.=C' = 2C.Now, = 1 q² = U1 2 C C' U2 1 q² C 2 C' = 2C (∵ C' = 2C) C U2 = U 2 Correct Option: A
In Ist case when capacitor C connected with battery charged with the energy,
U1 = U (stored energy on capacitor).
In IInd case, after disconnection of battery similar capacitor is connected in parallel with Ist capacitor then Ceq.=C' = 2C.Now, = 1 q² = U1 2 C C' U2 1 q² C 2 C' = 2C (∵ C' = 2C) C U2 = U 2
- Energy stored in a capacitor is
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Energy stored in capacitor
= 1 CV² = 1 Q y² (Q = CV) 2 2 V = 1 QV 2 Correct Option: A
Energy stored in capacitor
= 1 CV² = 1 Q y² (Q = CV) 2 2 V = 1 QV 2
- The capacity of a parallel plate condenser is 10 µF when the distance between its plates is 8 cm. If the distance between the plates is reduced to 4 cm then the capacity of this parallel plate condenser will be
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C = 10 µF
d = 8 cm
C' = ?
d' = 4 cmC = A∊0 ⇒ Cα 1 d d
If d is halved then C will be doubled.
Hence, C' = 2C = 2 × 10 µF = 20 µFCorrect Option: C
C = 10 µF
d = 8 cm
C' = ?
d' = 4 cmC = A∊0 ⇒ Cα 1 d d
If d is halved then C will be doubled.
Hence, C' = 2C = 2 × 10 µF = 20 µF