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Determine the electric field (magnitude and direction) at the point A (8.00 nm, 6.00 nm) caused by a particle located at the origin and carrying a charge of 7.00 μC .

User Nosredna
by
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1 Answer

5 votes

Answer:

E1 = 9.83 x
10^(20)
NC^(-1)

E2 = 1.748 x
10^(21)
NC^(-1)

Step-by-step explanation:

E1 = k Q/r2 = 8.99 x
10^(9) x 7 x
10^(-6) / 8 x
10^(-9) x 8 x
10^(-9) = 9.83 x
10^(20)
NC^(-1)

E2 = k Q/r2 = 8.99 x
10^(9) x 7 x
10^(-6) / 6 x
10^(-9) x 6 x
10^(-9) = 1.748 x
10^(21)
NC^(-1)

The direction of the electric field will be from E1 to E2...

User Uj Corb
by
7.0k points