The Earth is surrounded by an electric field on its surface that points toward the center of the Earth with an average magnitude of 150 N/C. If there is an electron that is released in vacuum at the Earth’s surface, what will the acceleration due to the Earth’s electric field be.
I got 2.6 × 1013 m/s2 am I right
acceleration = e E/m.
e = 1.6*10^-19 C
m = 9.1*10^-31 kg
E = 150 N/C
I do not agree with your answer.
To calculate the acceleration due to the Earth's electric field acting on an electron, we can use the equation:
a = F/m
Where:
a is the acceleration,
F is the force acting on the electron,
m is the mass of the electron.
In this case, we are given the magnitude of the electric field, which is 150 N/C. The force acting on the electron can be calculated using the equation:
F = qE
Where:
F is the force,
q is the charge of the electron (1.6 × 10^(-19) C),
E is the electric field.
Plugging in the values, we have:
F = (1.6 × 10^(-19) C) × (150 N/C)
Now, to find the acceleration, we divide the force by the mass of the electron:
a = [(1.6 × 10^(-19) C) × (150 N/C)] / (9.1 × 10^(-31) kg)
Evaluating the expression, we get:
a ≈ 2.6 × 10^14 m/s^2
So, it seems there might have been a calculation error. The correct answer for the acceleration due to the Earth's electric field acting on an electron is approximately 2.6 × 10^14 m/s^2.