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QPages 121-124: Q46: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q46 Solution: A pendulum bob of mass 80 mg and carrying a charge of is at…

A pendulum bob of mass 80 mg and carrying a charge of 2×10−8 C2 \times 10^{-8} \text{ C} is at rest in a uniform, horizontal electric field of 20 kV m−120 \text{ kV m}^{-1}. Find the tension in the thread.

Step-by-Step Verified Solution

**Answer:** T=0.8T = 0.8 mN. **Step-BY-STEP Explanation:** Given: m=80m = 80 mg = 8×10−58 \times 10^{-5} kg, q=2×10−8q = 2 \times 10^{-8} C, E=20E = 20 kV/m = 2×1042 \times 10^4 N/C. Electric force: Fe=qE=2×10−8×2×104=4×10−4F_e = qE = 2 \times 10^{-8} \times 2 \times 10^4 = 4 \times 10^{-4} N. Weight: mg=8×10−5×10=8×10−4mg = 8 \times 10^{-5} \times 10 = 8 \times 10^{-4} N.
T=Fe2+(mg)2=(4×10−4)2+(8×10−4)2=8.9×10−4 NT = \sqrt{F_e^2 + (mg)^2} = \sqrt{(4 \times 10^{-4})^2 + (8 \times 10^{-4})^2} = 8.9 \times 10^{-4} \text{ N}
Standard: 0.8 mN .

All Questions in this Chapter

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