Question:
When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then:
When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then:
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Battery connected: \(V\) constant \(\Rightarrow C\uparrow \Rightarrow Q\uparrow\).
Updated On: Apr 16, 2026
- charge on capacitor increases
- potential difference across the capacitor increases
- energy stored increases
- capacity remains the same
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The Correct Option is A
Solution and Explanation
Concept:
• Battery connected \(\Rightarrow\) voltage \(V\) remains constant
• Dielectric introduced \(\Rightarrow\) capacitance increases
Step 1:Capacitance \[ C' = kC \Rightarrow \text{increases} \]
Step 2:Charge \[ Q = CV \Rightarrow Q' = C'V = kCV \] \[ \Rightarrow Q \text{ increases} \]
Step 3:Energy \[ U = \frac{1}{2}CV^2 \Rightarrow U' = \frac{1}{2}kCV^2 \Rightarrow U \text{ increases} \]
Step 4:Potential difference \[ V = \text{constant} \Rightarrow \text{no change} \] Conclusion : (A)
• Battery connected \(\Rightarrow\) voltage \(V\) remains constant
• Dielectric introduced \(\Rightarrow\) capacitance increases
Step 1:Capacitance \[ C' = kC \Rightarrow \text{increases} \]
Step 2:Charge \[ Q = CV \Rightarrow Q' = C'V = kCV \] \[ \Rightarrow Q \text{ increases} \]
Step 3:Energy \[ U = \frac{1}{2}CV^2 \Rightarrow U' = \frac{1}{2}kCV^2 \Rightarrow U \text{ increases} \]
Step 4:Potential difference \[ V = \text{constant} \Rightarrow \text{no change} \] Conclusion : (A)
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