To find the magnitude and direction of the force exerted on a charge in an electric field, we can use the equation:

$\mathbf{F} = q \cdot \mathbf{E}$

where:

• $\mathbf{F}$ is the force,
• $q$ is the charge,
• $\mathbf{E}$ is the electric field.

Given:

• $q = 3.84 \, \mu\text{C} = 3.84 \times 10^{-6} \, \text{C}$,
• $\mathbf{E} = 155 \, \text{N/C}$ (directed due east).

Magnitude of the Force

$|\mathbf{F}| = q \cdot E = (3.84 \times 10^{-6} \, \text{C}) \times (155 \, \text{N/C}) = 5.952 \times 10^{-4} \, \text{N}$

Direction of the Force

The direction of the force will be the same as the direction of the electric field, which is due east.

Final Answer:

• Magnitude of the force: $5.952 \times 10^{-4} \, \text{N}$ (or $0.5952 \, \text{mN}$)
• Direction of the force: Due east.

Would you like more details on this calculation, or do you have any other questions?

Here are 5 related questions you might find interesting:

1. How would the force change if the charge were doubled?
2. What happens to the force if the electric field strength is halved?
3. How would the direction of the force change if the charge were negative?
4. What is the electric potential energy associated with this charge in the given electric field?
5. How does the force on a charge change if it is placed in a non-uniform electric field?

Tip: Remember, the direction of the force on a positive charge is always along the direction of the electric field, while for a negative charge, it's opposite to the electric field's direction.