An electric field of 4.04 x 10^5 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -10.4 ?C at this spot? Number Units
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The electric field \( E \) is given as \( 4.04 \times 10^{5} \, \mathrm{N/C} \), and the charge \( q \) is given as \(-10.4 \, \mu \mathrm{C}\). Show more…
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