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Electrostatic Potential and Capacitance

  1. A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?
    1. The energy stored in the capacitor decreases K times.
    2. The chance in energy stored is
      1
      CV²
      1
      - 1
      K
    3. The charge on the capacitor is not conserved.
    4. The potential difference between the plates decreases K times.
Correct Option: C

Capacitance of the capacitor, ​C = Q/V
After inserting the dielectric, new capacitance ​C1 = K.C ​
New potential difference ​V1 = V/K

Ui =
1
CV² =
(∵ Q = CV)
22C

Uf =
=
=
C²V²
=
ui
2f2KC2KCK

∆u = uf – ui =
1
CV²
1
- 1
2K

​As the capacitor is isolated, so change will remain conserved p.d. between two plates of the capacitor
L =
Q
=
V
KCK



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