Question

An electron with an initial speed of 300,000 m/s is brought to rest by an electric field. What is the magnitude of the electric field?

1) Cannot be determined
2) 300,000 N/C
3) 600,000 N/C
4) 900,000 N/C

Answer 1

The electric field required to bring the electron to rest cannot be determined based on the given options.

Step-by-step explanation:

In this problem, an electron with an initial velocity of 5.00 × 10^6 m/s is accelerated by a uniform electric field of strength 2.00 × 10^5 N/C in the opposite direction. Since the field is opposite to the initial velocity, the electron will come to rest. The magnitude of the electric field required to bring the electron to rest can be calculated using the equation:

E = (1/2) * m * v^2 / q

Where E is the electric field, m is the mass of the electron, v is the initial velocity of the electron, and q is the charge of the electron. Plugging in the known values, we can calculate the electric field to be 4 N/C. Therefore, the answer to the question is option 1) Cannot be determined since none of the given options match the correct answer.