Verified Solution • 2025–26 Session moderate

QPages 165-170: Q50: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q50 Solution: If the above capacitor is connected across a 6.0 V battery,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the induced charge formula, calculating it as $Q/K$ instead of the correct $Q(1 - 1/K)$, which leads to an incorrect net surface charge. Remember that the net charge on the conductor plate is the free charge $Q$ minus the magnitude of the induced dielectric charge $Q'$, resulting in $Q/K$.

If the above capacitor is connected across a 6.0 V battery, find a) the charge supplied by the battery, b) the induced charge on the dielectric and c) the net charge appearing on one of the coated surfaces.

Step-by-Step Verified Solution

**Answer:** (a) Q=8.5×10−9Q = 8.5 \times 10^{-9} C (b) Q′=6.4×10−9Q' = 6.4 \times 10^{-9} C (c) Qnet=2.1×10−9Q_{net} = 2.1 \times 10^{-9} C. **Step-by-Step Explanation:** (a) Q=CV=1.42×10−9×6=8.5×10−9Q = CV = 1.42 \times 10^{-9} \times 6 = 8.5 \times 10^{-9} C. (b) Q′=Q(1−1/K)=8.5×10−9×(1−1/4)=6.4×10−9Q' = Q(1 - 1/K) = 8.5 \times 10^{-9} \times (1 - 1/4) = 6.4 \times 10^{-9} C. (c) Net charge = Q−Q′=2.1×10−9Q - Q' = 2.1 \times 10^{-9} C .

All Questions in this Chapter

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