A positive charge of 3.50 µC is located in an electric field of intensity 3.00 N/C directed toward the south. Find the magnitude and the direction of the force on the test charge. (remember E = Fon q' / q', and 1 µC = 1.00 x 10-6 C)

a.
8.6 x 105 toward the south
c.
1.2 x 106 toward the south
b.
1.05 x 10-5 toward the south
d.
10.5 toward the sout

its b

5f

To find the force on a test charge in an electric field, you can use the equation F = Eq, where F is the force, E is the electric field intensity, and q is the charge.

Given:
Charge of the test charge, q' = 3.50 µC = 3.50 x 10^-6 C
Electric field intensity, E = 3.00 N/C

Using the equation F = Eq, we can substitute the given values:
F = (3.00 N/C) * (3.50 x 10^-6 C)

F = 1.05 x 10^-5 N

So, the magnitude of the force on the test charge is 1.05 x 10^-5 N.

The force is directed toward the south, which is specified in the question.

Therefore, the answer is b. 1.05 x 10^-5 toward the south.

To find the magnitude and direction of the force on the test charge, we can use the formula:

F = q * E

Where:
F is the force on the test charge
q is the charge of the test charge
E is the electric field intensity

Given:
q = 3.50 µC = 3.50 x 10^(-6) C
E = 3.00 N/C (directed toward the south)

Substituting the values into the formula, we get:

F = (3.50 x 10^(-6) C) * (3.00 N/C)

F = 10.5 x 10^(-6) N

Now, let's convert the force to scientific notation:

F = 1.05 x 10^(-5) N

Therefore, the magnitude of the force is 1.05 x 10^(-5) N.

Since the electric field is directed toward the south, the force on the test charge will also be directed toward the south.

Hence, the correct answer is option b. 1.05 x 10^(-5) N toward the south.