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Calculate the value of two equal charges if they repel one another with a force of 0.1 N when situated 50 cm apart in a vacuum.

a) 1.7μC
b) 2.8 x 1⁻⁵μC
c) 3.2μC
d) 5 x ⁻³μC

1 Answer

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Final answer:

The value of each of the equal charges that repel each other with a force of 0.1 N when situated 50 cm apart in a vacuum is approximately 1.7 μC. Hence, option (a) is correct.

Step-by-step explanation:

To calculate the value of two equal charges that repel each other with a known force, we utilize Coulomb's law, which is represented by the following equation:

F = k \cdot \fracq1 \cdot q2{r^2}

Where:

  • F is the electrostatic force between the charges (in newtons)
  • k is Coulomb's constant (8.9875 \times 10^9 N m^2/C^2)
  • q1 and q2 are the magnitudes of the two charges (in coulombs)
  • r is the distance between the charges (in meters)

Given that the two charges are equal, we can say q1 = q2 = q.

Plugging in the known values:

0.1 N = (8.9875 \times 10^9 N m^2/C^2) \cdot \frac{q^2}{(0.50 m)^2}

Now we solve for q:

q^2 = \frac{0.1 N \cdot (0.50 m)^2}{8.9875 \times 10^9 N m^2/C^2}

q^2 = \frac{0.025 N m^2}{8.9875 \times 10^9 N m^2/C^2}

q^2 = 2.78 \times 10^{-12} C^2

q = \sqrt{2.78 \times 10^{-12} C^2}

q = 1.67 \times 10^{-6} C

q = 1.67 μC

Hence, the value of each charge is approximately 1.7 μC, which corresponds to option (a).

User Niriel
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