The force experienced by a charge in an electric field $\mathbf{E}$ is given by:

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

Where:

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

First, we calculate the electric field using the given force and charge of $+5.0 \, \text{mC}$:

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

Given:

• $\mathbf{F} = 0.64 \, \text{N}$ (in the positive y direction),
• $q = 5.0 \, \text{mC} = 5.0 \times 10^{-3} \, \text{C}$.

Thus:

$\mathbf{E} = \frac{0.64 \, \text{N}}{5.0 \times 10^{-3} \, \text{C}} = 128 \, \text{N/C}$

Now, when the charge is replaced with a $-2.7 \, \text{mC}$ charge, the force will be:

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

Where:

• $q = -2.7 \, \text{mC} = -2.7 \times 10^{-3} \, \text{C}$,
• $\mathbf{E} = 128 \, \text{N/C}$.

Thus:

$\mathbf{F} = (-2.7 \times 10^{-3} \, \text{C}) \times 128 \, \text{N/C} = -0.3456 \, \text{N}$

The negative sign indicates the force is in the opposite direction of the electric field. Since the electric field was in the positive y direction, the force on the $-2.7 \, \text{mC}$ charge will be:

$\mathbf{F} = 0.3456 \, \text{N} \, \text{in the negative y direction}.$

Summary:

If the charge is replaced with a $-2.7 \, \text{mC}$ charge, it will experience a force of $0.3456 \, \text{N}$ in the negative y direction.

Do you need more details or have any questions?

Here are 5 related questions to expand on this topic:

1. How is the direction of force determined in an electric field?
2. What is the relationship between electric field strength and the force on a charge?
3. How would the force change if the charge were doubled?
4. What happens to the force if the charge is positioned in a uniform vs. non-uniform electric field?
5. How does the electric field between two point charges behave?

Tip: The sign of a charge affects the direction of the force but not its magnitude if the electric field is constant.