Verified Solution • 2025–26 Session moderate

QPages 121-124: Q36: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q36 Solution: A point charge produces an electric field of magnitude at a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the distance directly into the denominator without squaring it, using $E= rac{kQ}{r}$ instead of the correct $E= rac{kQ}{r^{2}}$. Always square the distance (and convert cm to m) before computing $Q$; otherwise the charge will be off by a factor of $r$.

A point charge produces an electric field of magnitude 5.0 N C−15.0 \text{ N C}^{-1} at a distance of 40 cm from it. What is the magnitude of the charge?

Step-by-Step Verified Solution

**Answer:** Q=8.9×10−11Q = 8.9 \times 10^{-11} C. **Step-BY-STEP Explanation:** Given: E=5.0E = 5.0 N/C, r=0.4r = 0.4 m.
E=kQr2⇒Q=Er2k=5.0×(0.4)29×109=8.9×10−11 CE = \frac{k Q}{r^2} \Rightarrow Q = \frac{E r^2}{k} = \frac{5.0 \times (0.4)^2}{9 \times 10^9} = 8.9 \times 10^{-11} \text{ C}

All Questions in this Chapter

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