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A point charge at the origin of a coordinate system produces the electric field E=(56,000 N/C) x on the x-axis at the location x=-0.65m. Determine the sign and magnitude of the charge.

Sagot :

hamzaahmeds

Answer:

q = 1.95 x 10⁻⁶ C = 1.95 μC

Explanation:

The formula for the electric field produced by a point charge is given as follows:

[tex]E = \frac{kq}{r^2}\\[/tex]

where,

E = Electric Field strength = 56000 N/C

k = Colomb's Constant = 9 x 10⁹ Nm²/C²

q = magnitude of charge = ?

r = distance = 0.65 m

Therefore,

[tex]56000\ N/C = \frac{(9\ x\ 10^9\ Nm^2/C^2)q}{(0.56\ m)^2}\\\\q = \frac{(56000\ N/C)(0.56\ m)^2}{9\ x\ 10^9\ Nm^/C^2}\\\\[/tex]

q = 1.95 x 10⁻⁶ C = 1.95 μC

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