Verified Solution • 2025–26 Session moderate

QPage 163: Q1: Capacitors Solutions — HC Verma: Concepts of Physics - Volume 2

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistake the "charge on a capacitor" for the total charge placed on the plates (e.g., $Q_1$ or $Q_2$) instead of the equal and opposite charge residing on the facing surfaces; the excess charge spreads over the outer surfaces, so the stored charge is the average difference $\frac{Q_1-Q_2}{2}$. Remember to separate net plate charge from the inner‑surface charge that determines the capacitor's voltage.

Suppose a charge +Q1+Q_1 is given to the positive plate and a charge −Q2-Q_2 to the negative plate of a capacitor. What is the "charge on the capacitor"?

Step-by-Step Verified Solution

**Answer:** The charge on the capacitor is Q1−Q22\frac{Q_1 - Q_2}{2}. **Step-by-Step Explanation:** Let the charge on the inner facing surfaces be qq and −q-q. Consider a Gaussian surface inside the positive plate. The electric field inside a conductor is zero. By Gauss's law and charge conservation, the net charge on the capacitor is Q1−Q22\frac{Q_1 - Q_2}{2} .

All Questions in this Chapter

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