Question:

An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 249 m/s^2 .

Part A:

What is the magnitude of the electric field?

Answer: E=_____ N/C

Part B:

What is the direction of the electric field? (Choose one)

- to the west
- to the south
- to the north
- to the east

Please Help.

Part A:

To calculate the magnitude of the electric field, we can use Newton's second law of motion. The force on the electron in the electric field is given by F = m*a, where m is the mass of the electron and a is the acceleration. Since the electron is in the uniform electric field, the electric force is given by F = q*E, where q is the charge of the electron and E is the electric field strength.

Now, we can equate the two equations for force:

q*E = m*a.

To find the magnitude of the electric field, we rearrange the equation to solve for E:

E = (m*a) / q.

Since the provided information does not include the values for mass and charge of the electron, we cannot determine the exact magnitude of the electric field. However, if you have those values, you can substitute them into the equation and calculate E.

Part B:

Without knowing the specific orientation of the coordinate system or the direction of the electric field, we cannot determine the exact direction of the electric field from the given information. Therefore, none of the provided options (west, south, north, east) can be chosen as the definitive answer.

To find the magnitude of the electric field, we can use the equation:

F = q * E

where F is the force experienced by the electron, q is the charge of the electron, and E is the electric field.

Since the electron is accelerating, we can use Newton's second law:

F = m * a

where m is the mass of the electron and a is its acceleration.

Now, we need to find the charge of an electron. The charge of an electron is approximately -1.6 x 10^-19 C.

We can rearrange the first equation to solve for the electric field:

E = F / q

Since F = m * a, we can substitute it in:

E = (m * a) / q

The mass of an electron is approximately 9.11 x 10^-31 kg.

Now we have all the values we need to calculate the electric field:

E = (9.11 x 10^-31 kg * 249 m/s^2) / (-1.6 x 10^-19 C)

Simplifying the calculations, we get:

E ≈ -1.44 x 10^9 N/C

The negative sign indicates that the electric field is directed opposite to the acceleration of the electron.

Therefore, the magnitude of the electric field is approximately 1.44 x 10^9 N/C, and the direction is to the south.

Answer:

Part A: E = 1.44 x 10^9 N/C
Part B: The direction of the electric field is to the south.

To find the magnitude of the electric field (Part A), we can use Newton's second law, F = qE, where F is the force experienced by the electron, q is the charge of the electron, and E is the electric field strength.

In this case, the force experienced by the electron is the force of the electric field, given by F = m*a, where m is the mass of the electron and a is the acceleration.

Since we know the acceleration (a = 249 m/s^2) and the mass of an electron (m = 9.11 x 10^-31 kg), we can substitute these values into F = m*a to find the force (F).

Next, we can rearrange the equation F = qE to solve for E:

E = F/q

Since we now have the value of F and the charge of an electron (q = 1.60 x 10^-19 C), we can substitute these values into the equation to find E.

Evaluating the equation will give you the magnitude of the electric field (E).

For Part B, to determine the direction of the electric field, we can use the fact that the electron is accelerating to the north. Since negative charges (like electrons) accelerate in the direction opposite to the electric field, we can conclude that the electric field must be to the south. Therefore, the direction of the electric field is "to the south" (Part B answer 2).

Remember to use the appropriate units for all calculations.