Question Solved1 Answer An electric field of \( 250 \mathrm{~N} / \mathrm{C} \), pointing due east, exerts a force on a \( 3.5 \) \( \mu \mathrm{C} \) charge. Determine the magnitude and direction of the force. \( 2.63 \times 10^{-4} N \), East \( 8.75 \times 10^{-4} N \), East \( 8.75 \times 10^{-2} N \), West \( 8.75 \times 10^{-6} N \), North \( 4.38 \times 10^{-4} N \), West \( An electric field of \( 250 \mathrm{~N} / \mathrm{C} \), pointing due east, exerts a force on a \( 3.5 \) \( \mu \mathrm{C} \) charge. Determine the magnitude and direction of the force. \( 2.63 \times 10^{-4} N \), East \( 8.75 \times 10^{-4} N \), East \( 8.75 \times 10^{-2} N \), West \( 8.75 \times 10^{-6} N \), North \( 4.38 \times 10^{-4} N \), West \( 14 \times 10^{-9} N \), West\( 71.4 \times 10^{-6} N \), South \( 3.06 \times 10^{-4} N \), East

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Transcribed Image Text: An electric field of \( 250 \mathrm{~N} / \mathrm{C} \), pointing due east, exerts a force on a \( 3.5 \) \( \mu \mathrm{C} \) charge. Determine the magnitude and direction of the force. \( 2.63 \times 10^{-4} N \), East \( 8.75 \times 10^{-4} N \), East \( 8.75 \times 10^{-2} N \), West \( 8.75 \times 10^{-6} N \), North \( 4.38 \times 10^{-4} N \), West \( 14 \times 10^{-9} N \), West \( 71.4 \times 10^{-6} N \), South \( 3.06 \times 10^{-4} N \), East
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Transcribed Image Text: An electric field of \( 250 \mathrm{~N} / \mathrm{C} \), pointing due east, exerts a force on a \( 3.5 \) \( \mu \mathrm{C} \) charge. Determine the magnitude and direction of the force. \( 2.63 \times 10^{-4} N \), East \( 8.75 \times 10^{-4} N \), East \( 8.75 \times 10^{-2} N \), West \( 8.75 \times 10^{-6} N \), North \( 4.38 \times 10^{-4} N \), West \( 14 \times 10^{-9} N \), West \( 71.4 \times 10^{-6} N \), South \( 3.06 \times 10^{-4} N \), East
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GivenElectric jiecd, epsilon=250N//C charge, q=3.5.11c.we know tixat{:[epsilon=(f)/(q)],[F=epsilon.q],[F=250 xx3.5 xx10^(-6)(N)/(F)xxC],[F=8.75 xx1 ... See the full answer