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MCQMediumJEE 2023Coulomb's Law & Superposition Principle

JEE Physics 2023 Question with Solution

Three point charges q=−2qq = -2q and 2q2q are placed on x-axis at a distance x=0x = 0, x=23Rx = \frac{2}{3} R and x=Rx = R respectively from origin as shown. If q=2×10−4 Cq = 2 \times 10^{-4} \, C and R=2 cmR = 2 \, cm, the magnitude of net force experienced by the charge −2q-2q is _____ N.

  • A

    5440 N5440 \, N

  • B

    54400 N54400 \, N

  • C

    5400 N5400 \, N

  • D

    440 N440 \, N

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Charges qq, −2q-2q and 2q2q are placed on the x-axis. The solution uses the position of −2q-2q as x=34Rx = \frac{3}{4}R and evaluates the net force on it due to the other two charges. It also states k=9×109 N m2/C2k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2, q=2×10−6 Cq = 2 \times 10^{-6} \, \text{C} and R=2 cm=0.02 mR = 2 \, \text{cm} = 0.02 \, \text{m}.

Find: Magnitude of the net force on −2q-2q.

Force due to qq at x=0x = 0:

r1=34Rr_1 = \frac{3}{4}R FBA=kq⋅2qr12F_{BA} = k \frac{q \cdot 2q}{r_1^2} FBA=kq⋅2q(34R)2=k8q29R2F_{BA} = k \frac{q \cdot 2q}{\left(\frac{3}{4}R\right)^2} = k \frac{8q^2}{9R^2}

Force due to 2q2q at x=Rx = R:

r2=R−34R=14Rr_2 = R - \frac{3}{4}R = \frac{1}{4}R FBC=k2q⋅2qr22F_{BC} = k \frac{2q \cdot 2q}{r_2^2} FBC=k8q2R2F_{BC} = k \frac{8q^2}{R^2}

Net force on −2q-2q:

FB=FBC−FBAF_B = F_{BC} - F_{BA} FB=k8q2R2−k8q29R2F_B = k \frac{8q^2}{R^2} - k \frac{8q^2}{9R^2} FB=k8q2R2(1−19)F_B = k \frac{8q^2}{R^2}\left(1 - \frac{1}{9}\right) FB=k8q2R2⋅89F_B = k \frac{8q^2}{R^2} \cdot \frac{8}{9}

Substituting the values used in the solution:

FB=9×109⋅8⋅(2×10−6)2(0.02)2⋅89F_B = \frac{9 \times 10^9 \cdot 8 \cdot (2 \times 10^{-6})^2}{(0.02)^2} \cdot \frac{8}{9} FB=9×109⋅8⋅4×10−124×10−4⋅89F_B = \frac{9 \times 10^9 \cdot 8 \cdot 4 \times 10^{-12}}{4 \times 10^{-4}} \cdot \frac{8}{9} FB=288×10−34×10−4⋅89F_B = \frac{288 \times 10^{-3}}{4 \times 10^{-4}} \cdot \frac{8}{9} FB=720⋅89=640 NF_B = 720 \cdot \frac{8}{9} = 640 \, \text{N}

the solution finally concludes that the net force is 5440 N5440 \, \text{N}. There is a clear discrepancy between the intermediate arithmetic shown, the given question values, and the final printed conclusion. Since the solution itself labels the correct answer as 54405440, the correct option is A.

Discrepancy Noted from Source

the content is internally inconsistent:

  • The question text places −2q-2q at x=23Rx = \frac{2}{3}R.
  • the solution computes using x=34Rx = \frac{3}{4}R.
  • The question text gives q=2×10−4 Cq = 2 \times 10^{-4} \, \text{C}.
  • the solution substitutes q=2×10−6 Cq = 2 \times 10^{-6} \, \text{C}.
  • The working shown gives 640 N640 \, \text{N}, but the solution heading and conclusion state 5440 N5440 \, \text{N}.

Because the instructions prioritize the solution, the extracted answer is A corresponding to 5440 N5440 \, \text{N}.

Common mistakes

  • Using the answer key without checking the solution working. Here the source has contradictions, so the solution must be treated.

  • Missing the direction of electrostatic forces on −2q-2q. One force is attractive and the other is repulsive, so the net force is found by vector subtraction, not direct addition.

  • Using inconsistent source values for qq or the position of −2q-2q. the question and the solution do not match, so mixing them blindly produces a wrong magnitude.

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