Atest charge is placed at a distance of 2.0 × 10-1 meters from a charge of 5.4 × 10-8 coulombs. what is the electric field at the test charge? (k = 9.0 × 109 newton·meter2/coulomb2)

a. 1.0 x 10^-2 newtons/coulomb
b. 1.2 x 10^4 newtons/coulomb
c. 2.5 x 10^4 newtons/coulomb
d. 3.5 x 10^-1 newtons/coulomb
e. 5.4 x 10^-8 newtons/coulomb

The electric field E of a charge is defined as E=F/Q where F is the Coulomb force and Q is the test charge. 

E=(1/Q)*k*(q*Q)/r², where k=9*10^9 N*m²/C², q is the point charge, Q is the test charge and r is the distance between the charges.

So E=(k*q)/r² 

When we input the numbers we get that electric field E of a point chage q is:

E=(9*10^9)*(5.4*10^-8)/0.2²=486/0.04=12150 N/C.
This is roughly E=12000 N/C =1.2*10^4 N/C
The correct answer is B. 

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