Question

5. A positive test charge of 5.0×10-6 C is in an electric field that exerts a force of 2.0×10-4 N to the right on it. What is the magnitude and direction of the electric field at the location of the test charge?

6. A negative charge of 2.0×10-8 C experiences a force of 0.060 N to the right in an electric field. What are the field’s magnitude and direction at that location?

7. A positive charge of 3.0×10-7 C is located in a field of 27 N/C directed toward the south. What is the force acting on the charge?

Answer 1

Electric field's direction is away from positive charge and towards the negative Charge

And Electrostatic force is given by:

F = q*E

E = F/q

Part A

Force is in right direction and charge is positive So direction of electric field will be towards left.

|E| = |F/q|

|E| = 2*10^-4/(5*10^-6) = 40 N/C

Part B

Force is in right direction and charge is negative So direction of electric field will be towards right.

|E| = |F/q|

|E| = 0.060/(2*10^-8) = 3*10^6 N/C

Part C

Electric field is in south direction and charge is positie So direction of force will be towards north.

F = q*E

F = 3*10^-7*27

F = 8.1*10^-6 N

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