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An electric field of $10^{5} N / C$ points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of $+ 2 μ C$ and $- 5 μ C$ at this spot?

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$0.8 N$ (due east) and $0.9 N$ (due west)

$0.8 N$ (due west) and $0.9 N$ (due east)

$0.2 N$ (due east) and $0.5 N$ (due west)

$0.2 N$ (due west) and $0.5 N$ (due east)

answer is D.

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Detailed Solution

Force on $+ 2 μ C = q E = (2 \times 10^{- 6}) (10^{5})$

$= 0.2 N$ (due west)

Force on $- 5 μ C = (5 \times 10^{- 6}) (10^{5})$

$= 0.5 N$ (due east)

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