Verified Solution • 2025–26 Session moderate

QPage 119: Q1: Electric Field Potential Solutions — HC Verma: Concepts of Physics - Volume 2

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often conflate magnitude with algebraic sign, thinking that because the electron’s charge is negative it “has $3.2\times10^{-19}\,$C less charge” than the proton; remember charge is a signed quantity, so the difference $q_{p}-q_{e}=+1.6-(-1.6)=3.2\times10^{-19}\,$C reflects subtraction of opposite signs, not that the electron’s magnitude is smaller.

The charge on a proton is +1.6×10−19 C+1.6 \times 10^{-19} \text{ C} and that on an electron is −1.6×10−19 C-1.6 \times 10^{-19} \text{ C}. Does it mean that the electron has 3.2×10−19 C3.2 \times 10^{-19} \text{ C} less charge than the proton?

Step-by-Step Verified Solution

**Answer:** No, it is not correct to say that the electron has 3.2 _ 10___ C less charge than the proton. **Step-by-Step Explanation:** The charge on a proton is +1.6 _ 10___ C and that on an electron is _1.6 _ 10___ C. The magnitude of charge on both is the same (1.6 _ 10___ C); they differ only in sign. The electron does not have "less charge" in magnitude; it has the same magnitude of charge but opposite sign. The difference in their charges is 3.2 _ 10___ C, but this is not the same as saying the electron has "less charge" .

All Questions in this Chapter

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