Question #295538

three point charges, which are equally separated by a distance 20 cm, are arranged along the y-axis in a vacuum. The topmost charge bears a charge of – 4.0μC, the middle charge has a charge of +3.0 μC, and the bottom one carries a 7.0 μC charge. What is the magnitude and direction of the net electrostatic force that the middle charge experiences?



1
Expert's answer
2022-02-09T08:03:53-0500

Answer

Charge

Q=-4μC\mu C

Q=3μCQ'=3\mu C

Q=7μCQ''=7\mu C

So all are seperated by d=0.2m

Net electrostatic force that the middle charge

F=KQQd2+KQ"Qd2F=\frac{KQQ'}{d^2}+\frac{KQ"Q'}{d^2}

So putting all values

F=9×109×4×1063×106(0.2)2+9×109×3×106×7×1060.2×0.2F=\frac{9\times10^{9}\times4\times10^{-6} 3\times10^{-6} }{(0.2)^2}+\frac{9\times10^{9}\times 3\times10^{-6} \times7\times10^{-6} }{0.2\times0.2}

So force is


F=7.425NF=7.425N



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