Physics Electrostatics Potential & Capacitance Properties of Charge and Coulomb's Law Subjective Type
Published on: September 11, 2026

Two point charges q 1 = 2 × 10 –3 C and q 2 = –3 × 10 –6 C are separated by a distance x = 10 cm. Find the magnitude and nature of the force between the two charges.

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The correct answer is:
A
Step 1: Use Coulomb's Law to calculate the force between the two charges. Coulomb's Law states that the force (F) between two point charges is given by:
$F = k \frac{|q_1 q_2|}{r^2}$
where $k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2$, $q_1 = 2 \times 10^{-3} \text{ C}$, $q_2 = -3 \times 10^{-6} \text{ C}$, and $r = 0.1 \text{ m}$ (which is 10 cm).

Step 2: Insert the values into the formula:
$F = 8.99 \times 10^9 \frac{|(2 \times 10^{-3})(-3 \times 10^{-6})|}{(0.1)^2}$

Step 3: Calculate the numerator:
$|q_1 q_2| = |(2 \times 10^{-3})(-3 \times 10^{-6})| = 6 \times 10^{-9} \text{ C}^2$

Step 4: Calculate the denominator:
$(0.1)^2 = 0.01 \text{ m}^2$

Step 5: Substitute back into the formula:
$F = 8.99 \times 10^9 \frac{6 \times 10^{-9}}{0.01}$

Step 6: Simplify:
$F = 8.99 \times 10^9 \times 6 \times 10^{-7} = 5.394 \text{ N}$

Step 7: Determine the nature of the force. Since $q_1$ is positive and $q_2$ is negative, the force is attractive.

Therefore, the magnitude of the force is 5.394 N, and the nature of the force is attractive.

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