Verified Solution • 2025–26 Session moderate

QPages 121-124: Q10: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q10 Solution: Suppose all the electrons of 100 g water are lumped together to …

Suppose all the electrons of 100 g water are lumped together to form a negatively-charged particle and all the nuclei are lumped together to form a positively-charged particle. If these two particles are placed 10.0 cm away from each other, find the force of attraction between them. Compare it with your weight.

Step-by-Step Verified Solution

**Answer:** Force = 2.56 _ 10__ N, which is about 10__ times the weight of a person. **Step-BY-STEP Explanation:** From Q9: total electron charge q1=−5.34×106q_1 = -5.34 \times 10^6 C. Total proton charge q2=+5.34×106q_2 = +5.34 \times 10^6 C. Distance r=0.1r = 0.1 m.
F=kq1q2r2=9×109×(5.34×106)20.01=2.56×1025 NF = \frac{k q_1 q_2}{r^2} = \frac{9 \times 10^9 \times (5.34 \times 10^6)^2}{0.01} = 2.56 \times 10^{25} \text{ N}
Weight of 50 kg person = 500 N. Ratio � 10__ .

All Questions in this Chapter

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