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Example · Example 4

Q.State the principle of superposition of electric forces. Two charges of +2 μC+2\ \mu\text{C} each are fixed at x=0x=0 and x=0.2 mx=0.2\ \text{m} on the xx-axis. Find the magnitude and direction of the net force on a third charge +1 μC+1\ \mu\text{C} placed at x=0.5 mx=0.5\ \text{m}.

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Superposition principle: the net force on a charge due to several others is the vector sum of the forces each other charge would exert alone.

Charge q0=+1 μCq_0=+1\ \mu\text{C} at x=0.5 mx=0.5\ \text{m}; source charges q1=+2 μCq_1=+2\ \mu\text{C} at x=0x=0 (distance r1=0.5 mr_1=0.5\ \text{m}) and q2=+2 μCq_2=+2\ \mu\text{C} at x=0.2 mx=0.2\ \text{m} (distance r2=0.3 mr_2=0.3\ \text{m}).

F1=kq1q0r12=(9×109)(2×10−6)(1×10−6)(0.5)2=0.0180.25=0.072 NF_1 = \frac{kq_1q_0}{r_1^2} = \frac{(9\times10^9)(2\times10^{-6})(1\times10^{-6})}{(0.5)^2} = \frac{0.018}{0.25} = 0.072\ \text{N}

F2=kq2q0r22=(9×109)(2×10−6)(1×10−6)(0.3)2=0.0180.09≈0.200 NF_2 = \frac{kq_2q_0}{r_2^2} = \frac{(9\times10^9)(2\times10^{-6})(1\times10^{-6})}{(0.3)^2} = \frac{0.018}{0.09} \approx 0.200\ \text{N} …

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