An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s2. What are the magnitude and direction of the electric field.

E=(ma/q)

Mass of electron=9.109e-31kg
Charge of electron=-1.602e-19C
so
E=(9.109e-31kg*115m/s^2)/-1.602e-19C
E=-6.539e-10
formated answer
E=6.539e-10 N/C South

Assuming I did it correctly:

F=ma

mass = the mass of the electron = 9.11 * 10^-31 kg

a= 115 m/s

plug in your known variables
F will equal 1.05 * 10^-28 Newtons (N)

Next, F=qE

q equals the charge of the electron, or "point charge".
q=1.6 * 10^-19 C

Solve for E, so E=F/q

plug in your values, and E will equal 6.56 *10^-10 N/C

Well, the electric field must be pretty shocked to see an electron accelerating to the north like that! It's pulling all the stops to make that happen! The magnitude of the electric field can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the force is the charge of the electron (which is the elementary charge, e) times the electric field strength (E). We can also use the equation for acceleration, which is equal to the force divided by the mass. Rearranging these equations, we find that the electric field strength is equal to the acceleration divided by the charge of the electron. So, the magnitude of the electric field would be 115 m/s^2 (acceleration) divided by the elementary charge (1.6 x 10^-19 C). As for the direction, well, considering that the electron is accelerating to the north, the electric field must be directed towards the south. So, the magnitude of the electric field is 7.2 x 10^20 N/C and it's headed south! Just goes to show, electrons can make any field get a little shocked!

To find the magnitude and direction of the electric field, we can use the equation of motion for an object under constant acceleration:

a = F / m

where a is the acceleration, F is the force, and m is the mass of the object.

In this case, the only force acting on the electron is the electric force:

F = qE

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

Since the electron is accelerating in the north direction, the electric force must be pointing in the same direction. Therefore, we can write the equation as:

qE = ma

where q is the charge of the electron, m is the mass of the electron, a is the acceleration, and E is the electric field.

The charge of an electron is -1.6 x 10^-19 C, and its mass is 9.1 x 10^-31 kg. The acceleration is given as 115 m/s^2.

Substituting these values into the equation, we can solve for the electric field:

(-1.6 x 10^-19 C)E = (9.1 x 10^-31 kg)(115 m/s^2)

E = (9.1 x 10^-31 kg)(115 m/s^2) / (-1.6 x 10^-19 C)

E ≈ -1.35 x 10^12 N/C

The magnitude of the electric field is approximately 1.35 x 10^12 N/C, and the direction is north.

To determine the magnitude and direction of the electric field, we can use the equation for the force experienced by a charged particle in an electric field.

The force experienced by a charged particle in an electric field is given by the equation: F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength.

In this case, the charge of the electron is q = -1.6 x 10^-19 C (since electrons have a negative charge).

The acceleration of the electron is given by the equation: a = F/m, where m is the mass of the electron.

Rearranging the equation F = qE, we can solve for the electric field strength E:

E = F/q

From the given information, the acceleration of the electron is a = 115 m/s^2 and the charge of the electron is q = -1.6 x 10^-19 C. The mass of the electron is approximately 9.1 x 10^-31 kg.

Plugging these values into the equation, we have:

E = (q*a)/m

E = (-1.6 x 10^-19 C * 115 m/s^2) / (9.1 x 10^-31 kg)

Calculating this equation, we find that the magnitude of the electric field is approximately 2.04 x 10^12 N/C.

The direction of the electric field can be determined by the direction of the acceleration. In this case, since the electron is accelerating to the north, the direction of the electric field is also to the north.