A positive test charge of 6.00 µC is placed in an electric field, which exerts a force of 5.00 x 10-4 N on the test charge. What is the magnitude of the electric field strength at the location of the test charge? (remember E = Fon q' / q', and 1 µC = 1.00 x 10-6 C)
12kg
1.2 x 104 N/C
grbf
To find the magnitude of the electric field strength at the location of the test charge, we can use the formula:
E = F / q
Where:
E is the electric field strength (in N/C)
F is the force exerted by the electric field on the test charge (in N)
q is the magnitude of the test charge (in C)
In this case, we are given:
F = 5.00 x 10^(-4) N
q = 6.00 x 10^(-6) C
Plugging these values into the formula, we get:
E = (5.00 x 10^(-4) N) / (6.00 x 10^(-6) C)
To simplify the calculation, we can also divide the numerator and the denominator by 10^(-6):
E = (5.00 x 10^(-4) N) / (6.00 x 10^(-6) C)
= (5.00 / 6.00) x (10^(-4) / 10^(-6)) N/C
= 0.833 x 10^(2) N/C
Finally, we can express this result in scientific notation:
E = 8.33 x 10^(1) N/C
Therefore, the magnitude of the electric field strength at the location of the test charge is 8.33 N/C.