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 say C, any help would be greatly appreciated.
a. 0 m/s2
b. 2.4 × 10-17 m/s2
c. 9.8 m/s2
d. 150 m/s2
e. 2.6 × 1013 m/s2
And you are very definitely wrong. I will be happy to critique your thinking.
Please no more answer grazing.
i am try to solve this question but answer is different and it equal 16960 *10^-27
It's e.
To solve this question, we need to understand the relationship between electric field and the acceleration of charged particles. The formula for the acceleration due to an electric field is given by:
a = q * E / m
where:
a is the acceleration
q is the charge of the particle
E is the electric field
m is the mass of the particle
In this case, we have an electron (negative charge) and the electric field on Earth's surface is given as 150 N/C.
The charge of an electron is approximately -1.6 x 10^-19 C, and the mass of an electron is approximately 9.1 x 10^-31 kg.
Now, let's substitute the values into the formula:
a = (-1.6 x 10^-19 C) * (150 N/C) / (9.1 x 10^-31 kg)
Calculating this expression, we find:
a ≈ -2.63 x 10^13 m/s^2
Since the question asks for the magnitude of the acceleration, we take the absolute value of this result:
|a| ≈ 2.63 x 10^13 m/s^2
Therefore, the answer is e. 2.6 × 10^13 m/s^2.