A parallel plate capacitor, each with plate area and separation , is charged to a potential difference . The battery used to charge it is then disconnected. A dielectric slab of thickness and dielectric constant is now placed between the plates. What change, if any, will take place in:
(i) charge on the plates,
(ii) electric field intensity between the plates,
(iii) capacitance of the capacitor? Justify your answer in each case.
(i) charge on the plates,
(ii) electric field intensity between the plates,
(iii) capacitance of the capacitor? Justify your answer in each case.
Solution
The battery is disconnected, so the charge on the plates is isolated. (i) Charge remains unchanged, as there is no path for it to flow. (ii) The dielectric gets polarised and its bound charges set up a field opposing the original one, so the net field becomes (reduced). (iii) Since and decreases by the factor while is fixed, the capacitance increases to .
CBSE Class 12 Board · 2007 · Subjective