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QPages 121-124: Q35: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q35 Solution: Two particles A and B possessing charges of and of ,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often confuse the conditions for zero field and zero potential by treating the electric field as a scalar; remember that $E$ is a vector requiring directional cancellation (only possible outside the charges for opposite signs), while $V$ is a scalar allowing algebraic cancellation (possible between charges).

Two particles A and B possessing charges of +2.00×10−6 C+2.00 \times 10^{-6} \text{ C} and of −4.00×10−6 C-4.00 \times 10^{-6} \text{ C}, respectively, are held fixed at a separation of 20.0 cm. Locate the points s) on the line AB, where a) the electric field is zero b) the electric potential is zero.

Step-by-Step Verified Solution

**Answer:** (a) Electric field zero at 20 cm from the -4 _C charge (outside). (b) Potential zero at 20/3 cm from the +2 _C charge (between) and 20 cm from the +2 _C charge (outside). **Step-BY-STEP Explanation:** Given: q1=+2.00×10−6q_1 = +2.00 \times 10^{-6} C, q2=−4.00×10−6q_2 = -4.00 \times 10^{-6} C, separation = 20 cm. (a) For E=0E = 0: kq1x2=k∣q2∣(x+0.2)2\frac{k q_1}{x^2} = \frac{k |q_2|}{(x+0.2)^2}. Solving gives x=0.2x = 0.2 m = 20 cm from the negative charge. (b) For V=0V = 0: kq1x=kq20.2−x\frac{k q_1}{x} = \frac{k q_2}{0.2 - x} (between charges). Solving gives x=0.0667x = 0.0667 m = 6.67 cm from the positive charge .

All Questions in this Chapter

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