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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance3 Marks
Question

A parallel plate capacitor, each with plate area A and separation d, is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change, if any, will take place in.

  1.  Charge on the plates.
  2. Electric field intensity between the plates.
  3. Capacitance of the capacitor.
✓

Answer

  1. No change.

As the battery is disconnected.

  1. Decreases OR becomes $\frac{1}{\text{k}}$ times Due to polarisation of the dielectric.
  2. Increases OR becomes k times.

As the electric field, and therefore, the p.d., between the plates decreases.

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