The function of a dielectric in a capacitor is to
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- reduce the plate area of the capacitor.
- to decrease the capacitance.
- reduce the effective potential on plates.
- increase the effective potential on plates.
The Correct Option is C
Solution and Explanation
Keep $Q$ fixed. The field falls to $E_0/K$.
Step 2: Potential:
$V=Ed$ falls to $V_0/K$.
Step 3: Capacitance:
$C=Q/V$ rises. Option (C) describes the reduced potential.
Final Answer:
The induced charges oppose the original field. \[ \boxed{C} \]
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