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.