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QPages 121-124: Q58: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q58 Solution: An electric field exists in space. If the potential at the…

⚠️ Common Exam Traps & Conceptual Pitfalls

When computing $V(mathbf r)= -int_{mathbf 0}^{mathbf r}mathbf Ecdot dmathbf r$, students often drop the minus sign, treating the integral as positive. Remember that the potential decreases in the direction of the electric field, so for a constant field $V(mathbf r)= -mathbf Ecdotmathbf r$; keeping the negative sign avoids a sign error.

An electric field E⃗=20i^+30j^) N C−1\vec{E} = 20\hat{i} + 30\hat{j}) \text{ N C}^{-1} exists in space. If the potential at the origin is taken to be zero, find the potential at 2 m, 2 m).

Step-by-Step Verified Solution

**Answer:** V=−100V = -100 V. **STEP-BY-STEP Explanation:** E⃗=20i^+30j^\vec{E} = 20\hat{i} + 30\hat{j} N/C. V=−∫E⃗⋅dr⃗V = -\int \vec{E}\cdot d\vec{r}. At origin V=0V = 0. At (2, 2):
V=−(20×2+30×2)=−100 VV = -(20 \times 2 + 30 \times 2) = -100 \text{ V}

All Questions in this Chapter

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