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3 years ago

14

What magnitude charge creates a 32.08 N/C electric field at a point 2.84 m away?

1 answer:

sammy [17]3 years ago

5 0

Answer:

Q = 29.4 x 10⁻⁹ C.

Explanation:

Electric field due to a charge Q at distance d is given by coulomb law as follows

Electric field

E = k Q /d²

where for air k which is a constant is 9 x 10⁹

$E=\frac{k\times Q}{d^2}$

Given E = 32.08 , d = 2.84 m

Putting these values in the relation above, we have

$[tex]32.08=\frac{9\times 10^9\times Q}{(2.84)^2}$ [/tex]

Q = 29.4 x 10⁻⁹ C.

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