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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Electric Field and Potential

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Field and Potential

Two equal charges are situated at a separation of 1.0 m . What must...

Two equal charges are situated at a separation of 1.0 m1.0 \text{ m}. What must be the magnitude of the charges so that the force between them is equivalent to the weight of a 50 kg50 \text{ kg} person?

Options

  1. A2.3×10−4 C2.3 \times 10^{-4} \text{ C}
  2. B4.6×10−4 C4.6 \times 10^{-4} \text{ C}
  3. C1.2×10−4 C1.2 \times 10^{-4} \text{ C}
  4. D2.3×10−5 C2.3 \times 10^{-5} \text{ C}

Correct answer

A. 2.3×10−4 C2.3 \times 10^{-4} \text{ C}

Step-by-step solution

Weight of the person, W=mg=50×9.8=490 NW = mg = 50 \times 9.8 = 490 \text{ N}

Electrostatic force between the charges, F=14πϵ0q2r2F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r^2}

Substituting the given values:

F=9×109×q212=9×109q2F = \dfrac{9 \times 10^9 \times q^2}{1^2} = 9 \times 10^9 q^2

Equating the electrostatic force to the weight:

9×109q2=4909 \times 10^9 q^2 = 490

q2=4909×109=5.44×10−8q^2 = \dfrac{490}{9 \times 10^9} = 5.44 \times 10^{-8}

q=5.44×10−8≈2.33×10−4 Cq = \sqrt{5.44 \times 10^{-8}} \approx 2.33 \times 10^{-4} \text{ C}
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