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3 years ago

What is the electric field at a location <-0.5, -0.1, 0> m, due to a particle with charge +4 nC located at the origin?

Physics

1 answer:

Natasha2012 [34]3 years ago

7 0

Answer:

138.4625 N/C

Explanation:

We have that the charge +4 nC is at the origin (0,0,0)

We have to find the electric field at point (-0.5,-0.1,0)

We know that distance between two points is given by $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}=\sqrt{(0-(-0.5))^2+(0-(-0.1))^2+(0-0)^2}=0.5099m$

The electric field due to a point charge is given by $E=\frac{KQ}{R^2}=\frac{4\times 10^{-9}\times 9\times 10^{9}}{0.5099^2}=138.4625N/C$

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Answer:

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AVprozaik [17]

Missing part in the exercise. Complete text:
"A hair dryer draws 14.5 A when plugged into a 120-V line. Assume direct current.

(a) What is its resistance? Express your answer to two significant figures and include the appropriate units."

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Solution

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