A proton is initially at rest in empty space, when suddenly an electric field of magnitude 8.0 x 10 4 n/c and direction due east is turned on. what is the magnitude and direction of the initial acceleration of the proton? Repeat the problem, but with an electron.

To calculate the magnitude and direction of the initial acceleration of a proton or an electron in an electric field, we need to use the formula:

a = qE/m

where:
a = acceleration
q = charge of the particle
E = magnitude of the electric field
m = mass of the particle

Let's first calculate the acceleration for a proton:
q = charge of a proton = 1.6 x 10^(-19) C
E = magnitude of the electric field = 8.0 x 10^4 N/C (given)
m = mass of a proton = 1.67 x 10^(-27) kg

Plugging these values into the formula, we get:
a = (1.6 x 10^(-19) C x 8.0 x 10^4 N/C) / (1.67 x 10^(-27) kg)

Calculating this expression, we find:
a ≈ 9.58 x 10^6 m/s^2

The direction of the initial acceleration for a positive charge like a proton is the same as the direction of the electric field, which in this case is due east.

Now let's repeat the problem for an electron:
q = charge of an electron = -1.6 x 10^(-19) C (negative charge)
E = magnitude of the electric field = 8.0 x 10^4 N/C (given)
m = mass of an electron = 9.11 x 10^(-31) kg

Using the same formula, we can find the acceleration:
a = (-1.6 x 10^(-19) C x 8.0 x 10^4 N/C) / (9.11 x 10^(-31) kg)

Evaluating this expression, we get:
a ≈ -1.76 x 10^18 m/s^2

Since the charge of an electron is negative, the direction of the initial acceleration for an electron is opposite to the direction of the electric field, which in this case is due east.