When a test charge q_0 = 8 is placed at the origin, it experiences a force of 8

Question

When a test charge q_0 = 8 is placed at the origin, it experiences a force of 8 times 10^-4 N in the positive y direction (a) What is the electric field at the origin? 1e5 N/C j (b) What would be the force on a charge of -4 nC placed at the origin? -4e-4 N j (c) If this force is due to a charge on the y axis at y = 3 cm, what is that charge (magnitude and sign)? -2.7e - 10 nC A charge of 3 mu C is at the origin. What is the magnitude and direction of the electric field on the x axis at each of the following locations? (Remember that E_x is negative when E points in the negative x direction.) (a) x = 6 m 75 N/C i (b) x = -10 m N/C i

Explanation / Answer

Ans1)

We know F = qE

8e-4 = 8e-9*E

E = 1e5 V/m

Ans2)

F = qE = -4e-9*1e5 = -4e-4 j N

Ans 3)

F = kq1q2/d^2

4e-4 = 9e9*4e-9*q2/(0.03*0.03)

q2 = 4e-4*(0.03*0.03)/(36) = 1e-8 C

Ans 2)

E = kq/d^2

(a) E = 9e9*3e-6/(6*6) = 750 V/m i

b)E= 9e9*3e-6/(10*10) = 270V/m

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