Question
In which of the following two cases, more work will be done in increasing the separation between the plates of a charged capacitor and why?
(i) The charging battery remains connected to the capacitor.
(ii) The battery is removed after charging the capacitor.
In the same question, if battery is removed after charging the condenser and dielectric slab introduced how are all the five parameters affected?
Solution
Capacitance C is given by C==
On increasing the seperation between the plates of a charged capacitor, capacitance decreases.
i) When charging battery remains connected to the capacitor, potential remains constant and charge Q decreases.
ii) When battery is removed after charging the capacitor, charge Q remains constant but, potential V decreases.
Hence, more amount of work is required to be done in the second case.
iii) When, battery is removed after charging the capacitor and dielectric slab is introduced.
Capacitance increases as for any dielectric K>1.
Charge Q remains constant.
Potential difference-V= decreases as capacitance increases.
Electric field E= decreases.
Energy stored in the capacitor decreases.
On increasing the seperation between the plates of a charged capacitor, capacitance decreases.
i) When charging battery remains connected to the capacitor, potential remains constant and charge Q decreases.
ii) When battery is removed after charging the capacitor, charge Q remains constant but, potential V decreases.
Hence, more amount of work is required to be done in the second case.
iii) When, battery is removed after charging the capacitor and dielectric slab is introduced.
Capacitance increases as for any dielectric K>1.
Charge Q remains constant.
Potential difference-V= decreases as capacitance increases.
Electric field E= decreases.
Energy stored in the capacitor decreases.
