An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s^2.Determine the magnitude and direction of electric field.

To determine the magnitude and direction of the electric field, we can use the equation:

a = F / m

where:
a = acceleration of the electron
F = force experienced by the electron
m = mass of the electron

Since the electron is accelerating in a uniform electric field, the force experienced by the electron is due to the electric field. The force is given by:

F = q * E

where:
q = charge of the electron
E = electric field

The mass of an electron is approximately 9.11 x 10^(-31) kg, and the charge of an electron is -1.6 x 10^(-19) C (since it is negatively charged).

Substituting these values into the equations, we have:

a = F / m
=> a = (q * E) / m

Rearranging the equation, we can solve for the electric field:

E = (m * a) / q

Substituting the given values:

m = 9.11 x 10^(-31) kg
a = 115 m/s^2
q = -1.6 x 10^(-19) C

E = ((9.11 x 10^(-31) kg) * (115 m/s^2)) / (-1.6 x 10^(-19) C)

Calculating this expression, we find:

E ≈ -5.2 x 10^8 N/C

The negative sign indicates that the electric field is directed toward the south.

Therefore, the magnitude of the electric field is approximately 5.2 x 10^8 N/C and its direction is toward the south.

To determine the magnitude and direction of the electric field, we can use the equation that relates the electric field to the acceleration of a charged particle:

E = ma/q

Where:
E is the electric field,
m is the mass of the charged particle,
a is the acceleration of the charged particle, and
q is the charge of the charged particle.

In this case, we are dealing with an electron, which has a charge of -1.6 x 10^-19 Coulombs and a mass of 9.11 x 10^-31 kg.

Plugging in the values:

E = (9.11 x 10^-31 kg) x (115 m/s^2) / (-1.6 x 10^-19 C)

Calculating this expression will give us the magnitude of the electric field.