Answer to Question #262919 in Electricity and Magnetism for Moses

Question #262919

Two point charges separated by a distance of 30cm in air. One of the charges is positive and has a magnitude of 10 micro coulombs and the other is a negative charge of magnitude 15 micro coulombs. Calculate the force between the charges, state if it is force of attraction or repulsion


1
Expert's answer
2021-11-08T17:12:36-0500

q1=10μcq2=15μcd=0.30mF=kq1q2r2q_1=10\mu c\\q_2=15\mu c\\d=0.30m \\F=\frac{kq_1q_2}{r^2}


F=9×109×10×106×(15)×1060.302=15NF=\frac{9\times10^9 \times10\times10^{-6}\times(-15)\times10^{-6}}{0.30^2}=-15N

Negative sign indicated Attractions force


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