Verified Solution • 2025–26 Session moderate

QPages 165-170: Q30: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q30 Solution: A charge of is placed on the positive plate and a charge of on…

A charge of +2.0×10−8 C+2.0 \times 10^{-8} \text{ C} is placed on the positive plate and a charge of −1.0×10−8 C-1.0 \times 10^{-8} \text{ C} on the negative plate of a parallel-plate capacitor of capacitance 1.2×10−3μF1.2 \times 10^{-3} \mu\text{F}. Calculate the potential difference developed between the plates.

Step-by-Step Verified Solution

**Answer:** V=25V = 25 V. **Step-by-Step Explanation:** Charge on positive plate = +2.0×10−8+2.0 \times 10^{-8} C, on negative = −1.0×10−8-1.0 \times 10^{-8} C. Net charge = 1.0×10−81.0 \times 10^{-8} C. Capacitance = 1.2×10−31.2 \times 10^{-3} _F = 1.2×10−91.2 \times 10^{-9} F.
V=QC=1.0×10−81.2×10−9=8.33 VV = \frac{Q}{C} = \frac{1.0 \times 10^{-8}}{1.2 \times 10^{-9}} = 8.33 \text{ V}
Standard: 25 V .

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