A 4.40 μC charge is subject to a 6.00 mN force due to an electric field. What is the magnitude of the electric field at the location of the charge?
Question
A 4.40 μC charge is subject to a 6.00 mN force due to an electric field. What is the magnitude of the electric field at the location of the charge?
Solution
The magnitude of the electric field (E) can be calculated using the formula:
E = F/q
where: F is the force = 6.00 mN = 6.00 x 10^-3 N (since 1mN = 10^-3 N) q is the charge = 4.40 μC = 4.40 x 10^-6 C (since 1μC = 10^-6 C)
Substituting the values into the formula, we get:
E = (6.00 x 10^-3 N) / (4.40 x 10^-6 C)
Solving this gives:
E = 1.36 x 10^3 N/C
So, the magnitude of the electric field at the location of the charge is 1.36 x 10^3 N/C.
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