Let's solve the problem step by step.

(a) Finding the Magnitude and Direction of the Electric Field

The force $\vec{F}$ on a charge $q$ in an electric field $\vec{E}$ is given by:

$\vec{F} = q \vec{E}$

The magnitude of the electric field $\vec{E}$ is:

$E = \frac{F}{q}$

Given:

• Force on the electron, $F = 4.80 \times 10^{-17} \, \text{N}$ (westward)
• Charge of an electron, $q_e = -1.60 \times 10^{-19} \, \text{C}$

Since the force is westward and the charge of the electron is negative, the electric field direction must be opposite to the force direction, i.e., eastward.

Now, calculate the magnitude of the electric field:

$E = \frac{4.80 \times 10^{-17} \, \text{N}}{1.60 \times 10^{-19} \, \text{C}} = 3.00 \times 10^2 \, \text{N/C}$

So, the magnitude of the electric field is $\mathbf{300 \, N/C}$ and its direction is eastward.

(b) Force on a Proton in the Same Electric Field

For a proton in the same electric field, we use the same formula:

$\vec{F} = q_p \vec{E}$

Given:

• Charge of a proton, $q_p = +1.60 \times 10^{-19} \, \text{C}$
• Electric field, $E = 300 \, \text{N/C}$ (eastward)

The magnitude of the force is:

$F = q_p \times E = 1.60 \times 10^{-19} \, \text{C} \times 300 \, \text{N/C} = 4.80 \times 10^{-17} \, \text{N}$

Since the proton has a positive charge, the force direction is the same as the electric field direction, which is eastward.

Summary:

• (a) The magnitude of the electric field is $\mathbf{300 \, N/C}$, and its direction is eastward.
• (b) The magnitude of the force on a proton is $\mathbf{4.80 \times 10^{-17} \, N}$, and its direction is eastward.

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. What happens if the direction of the electric field is reversed?
2. How would the forces change if the charge values were doubled?
3. What is the electric field if the force on the electron were $1.60 \times 10^{-17}$ N?
4. How does the electric field influence the acceleration of the proton?
5. What would be the force on an alpha particle (which has twice the charge of a proton) in this field?

Tip: The direction of the electric field is always taken as the direction of the force on a positive test charge.