An electron experiences a force of 12 mN in the positive x-direction in an electric field. What is the magnitude and direction of the electric field?
E field = Force/(electron charge)
Divide 12*10^-3 N by the electron charge in Coulombs.
You may need to look that up.
1 Newton per Coulomb is the same as 1 Volt per meter.
To find the magnitude and direction of the electric field, we can use the equation:
Electric force (F) = charge of electron (q) * electric field (E)
Given that the electric force acting on the electron is 12 mN (millinewtons) in the positive x-direction, and the charge of an electron (q) is a fundamental constant equal to 1.6 × 10^(-19) Coulombs, we can rearrange the equation to solve for the electric field (E):
E = F / q
Plugging in the known values:
E = 12 mN / 1.6 × 10^(-19) C
To divide millinewtons by Coulombs effectively, we need to convert the mN to Newtons (N). Since 1 N = 1,000,000 mN, we can convert 12 mN to Newtons by dividing it by 1,000,000:
E = 12 × 10^(-6) N / 1.6 × 10^(-19) C
Now, we can simplify the expression by dividing the numerator and denominator by 10^(-19):
E = (12 / 1.6) × (10^(-6) / 10^(-19)) N/C
E = 7.5 × 10^13 N/C
Thus, the magnitude of the electric field is 7.5 × 10^13 N/C.
Since the electric force is directed in the positive x-direction, the electric field must also be in the positive x-direction. Therefore, the direction of the electric field is the same as the direction of the force acting on the electron, which is the positive x-direction.