Find the electric field acting on a 2.0 C charge of an electrostatic force of 10,500 N acts on the particle.
Answer
Force on charge F=10500N
Charge q=2C
So electric field is given by
E=Fq=105002=5250N/CE=\frac{F}{q}\\=\frac{10500}{2}\\=5250N/CE=qF=210500=5250N/C
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