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Numerical · Q18

Q.Two point charges +2 μC+2\ \mu\text{C} and +3 μC+3\ \mu\text{C} are placed 0.3 m0.3\ \text{m} apart in air. Calculate the force between them.

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✓ Free question

Given q1=+2 μC=2×10−6 Cq_1=+2\ \mu\text{C}=2\times10^{-6}\ \text{C}, q2=+3 μC=3×10−6 Cq_2=+3\ \mu\text{C}=3\times10^{-6}\ \text{C}, r=0.3 mr=0.3\ \text{m}.

F=kq1q2r2=(9×109)(2×10−6)(3×10−6)(0.3)2F = \frac{kq_1q_2}{r^2} = \frac{(9\times10^9)(2\times10^{-6})(3\times10^{-6})}{(0.3)^2}

Numerator: 9×109×2×10−6=1.8×1049\times10^9\times2\times10^{-6}=1.8\times10^4; then 1.8×104×3×10−6=5.4×10−21.8\times10^4\times3\times10^{-6}=5.4\times10^{-2}.

Denominator: (0.3)2=0.09(0.3)^2=0.09.

F=5.4×10−20.09=0.6 NF = \frac{5.4\times10^{-2}}{0.09} = 0.6\ \text{N}

Since both charges are positive, the force is repulsive.

✓Final answer

F=0.6 NF=0.6\ \text{N}, repulsive.

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