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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 9 of 37
Marks: +1, -0
Explain what would happen if in the capacitor given in above question, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates.
(a) While the voltage supply remained connected.
(b) After the supply was disconnected.
Solution:  
C = KC0KC_0 = 6 × 18 = 108 pF
(a) If the voltage supply remained connected, then the potential difference across the capacitor will remain the same i.e. V = 100 V and hence charge on the capacitor becomes
Q = CV = 108 × 10−1210^{-12} × 100 or Q = 1.08 × 10−810^{-8} C
(b) If the voltage supply was disconnected, then charge on the capacitor remains the same
i.e., Q = 1.8 × 10−910^{-9} C
and hence potential difference across the capacitor becomes
V = QC\frac{Q}{C} = 1.8×10−9108×10−12F\frac{1.8 \times 10^{-9}}{108 \times 10^{-12} \mathrm{F}} or V = 16.6 V
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