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Suppose an electric field with a magnitude of 2.8 E 4 N/C at a certain location, points due south. If a charge of -4.0 E -6 C is brought to the location, what is the electrostatic force that acts on this
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Mar 25, 2019
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Suppose an electric field with a magnitude of 2.8 E 4 N/C at a certain location, points due south. If a charge of -4.0 E -6 C is brought to the location, what is the electrostatic force that acts on this charge?
Physics
high-school
OJ Kwon
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OJ Kwon
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The correct answer is: F
=
-0.112N
(Negative sign indicates that the force direction is opposite to South: meaning North)
Step-by-step explanation:
Since,
Electric field = Force / UnitCharge
E = F / q
=> F = Eq
Where F = Electrostatic force = ?
E = Electric field = 2.8 * 10^4 N/C
q = Charge = -4 * 10^-6 C
Plug in the values:
F = (2.8 * 10^4) * (-4 * 10^-6)
F = -0.112N (Negative sign indicates that the force direction is opposite to South: meaning North)
Saurabh Bayani
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Mar 29, 2019
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Saurabh Bayani
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