Verified Solution • 2025–26 Session moderate

QPages 121-124: Q34: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q34 Solution: The electric force experienced by a charge of is . Find the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often reverse the relation $F=qE$ and compute $E=qF$ instead of $E=F/q$. Remember that the electric field is the force per unit charge, so divide the force by the charge’s magnitude, not multiply.

The electric force experienced by a charge of 1.0×10−6 C1.0 \times 10^{-6} \text{ C} is 1.5×10−3 N1.5 \times 10^{-3} \text{ N}. Find the magnitude of the electric field at the position of the charge.

Step-by-Step Verified Solution

**Answer:** E=1500E = 1500 N/C. **Step-BY-STEP Explanation:** Given: q=1.0×10−6q = 1.0 \times 10^{-6} C, F=1.5×10−3F = 1.5 \times 10^{-3} N.
E=Fq=1.5×10−31.0×10−6=1500 N/CE = \frac{F}{q} = \frac{1.5 \times 10^{-3}}{1.0 \times 10^{-6}} = 1500 \text{ N/C}

All Questions in this Chapter

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