When a dielectric slab is introduced in between the plates of a charged capacitor with battery connected across the plates, following are the observations. Choose which observation is incorrect?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Potential changes
The battery stays connected, so it keeps the potential difference between the plates fixed at its emf V. Introducing the dielectric cannot change this potential difference; only the charge, the capacitance and the energy adjust.
\( C = KC_0 \) (the capacitance increases K times)
\( Q = CV = KC_0V \) (more charge flows in from the battery)
\( \displaystyle U = \frac{1}{2}CV^{2} \) (increases as C increases at constant V)
So the capacity and the energy increase, while the potential difference does not change. The observation “Potential changes” is therefore the incorrect one.
NCERT Exemplar:
Sample Paper I – “… 10. The battery remains connected to a parallel plate capacitor and a dielectric slab is inserted between the plates. What will be effect on its (i) potential difference, (ii) capacity, (iii) electric field, and (iv) energy stored?”
Sample Paper I Solutions and Marking Scheme – “… 10. When battery remains connected, (i) potential difference V remains constant. … (ii) capacity C increases. … (iii) electric field will decrease. … (iv) energy stored \( \frac{1}{2}CV^2 \) increases as C increases.”