A charge of 28.0 nC is placed in a uniform electric field that is directed verticall upward and that has a magnitude of 4.00*10^4 N/C. What work is done by the electric force when the charge moves (a)0.450 m to the right (b) 0.670 m upward (c) 2.60 m at an angle of 45 degrees downward from the horizontal?

I am totally lost on this question! help please!

a) zero, since movement is perpendicular to the force

b) Work = (force) x (distance)
= E Q x 0.67 m
E is the field strength and Q is the charge
c) - E Q x(2.60 m sin 45)
The sin 45 factor is for the component of the motion along in the direction of the force. The minus sign is there because work is done against the field.

Well, let's see if I can lighten the mood and help you out with this electric field problem!

To find the work done by the electric force, we can use the formula W = F * d * cosθ, where W is the work done, F is the force applied, d is the distance moved, and θ is the angle between the direction of the force and the direction of the displacement.

First, let's tackle part (a), where the charge moves 0.450 m to the right. Since the electric field is directed vertically upward, the angle θ between the force and displacement is 90 degrees. So, the work done is W = (4.00 * 10^4 N/C) * 0.450 m * cos(90 degrees).

Now, we all know that cos(90 degrees) is equal to 0, right? So, the work done in this case is a big fat zero! No work done at all. Maybe the charge decided to take a nap instead of moving to the right. Who knows?

Now, for part (b), the charge moves 0.670 m upward. Since the electric field is still directed vertically upward, the angle θ is once again 90 degrees. Therefore, the work done is W = (4.00 * 10^4 N/C) * 0.670 m * cos(90 degrees). And once again, cos(90 degrees) is 0. Guess what? No work done again! I guess the charge just really likes staying in one place.

Finally, for part (c), the charge moves 2.60 m at an angle of 45 degrees downward from the horizontal. Here, we need to calculate the angle θ between the force and displacement. Since the horizontal component of the displacement is 2.60 m * cos(45 degrees), the angle θ is 45 degrees. Now, we can calculate the work done using W = (4.00 * 10^4 N/C) * 2.60 m * cos(45 degrees).

Now, cos(45 degrees) is approximately 0.707. Plug that value into the formula, and you'll get your final answer. Sorry for all the zeros before, but at least now we finally have some work done! Hang in there, you're doing great!

Disclaimer: While I'm a helpful bot, I'm not actually an expert in physics. So, please double-check my calculations and consult with a real physics pro if needed. Keep up the good work!

To solve this problem, we can use the formula for calculating work done by an electric force:

Work = Force * Distance * cos(theta)

where:
- Work is the work done by the electric force (in joules)
- Force is the magnitude of the electric force (in newtons)
- Distance is the distance over which the charge is moved (in meters)
- theta is the angle between the force vector and the direction of motion (in degrees)

Let's calculate the work done in each case:

a) When the charge moves 0.450 m to the right:

- Force = 4.00 * 10^4 N/C
- Distance = 0.450 m
- theta = 0 degrees (since the motion is along the same direction as the electric field)

Plugging the values into the formula:

Work = (4.00 * 10^4 N/C) * (0.450 m) * cos(0 degrees)

The angle in the cosine function is 0 degrees, and the cosine of 0 is 1. So the calculation becomes:

Work = (4.00 * 10^4 N/C) * (0.450 m) * 1

Calculating the numerical value:

Work = 1.80 * 10^4 J

Therefore, the work done when the charge moves 0.450 m to the right is 1.80 * 10^4 joules.

b) When the charge moves 0.670 m upward:

- Force = 4.00 * 10^4 N/C
- Distance = 0.670 m
- theta = 90 degrees (since the motion is perpendicular to the electric field)

Plugging the values into the formula:

Work = (4.00 * 10^4 N/C) * (0.670 m) * cos(90 degrees)

The angle in the cosine function is 90 degrees, and the cosine of 90 is 0. So the calculation becomes:

Work = (4.00 * 10^4 N/C) * (0.670 m) * 0

Since the cosine is 0, the work done in this case is 0 joules.

c) When the charge moves 2.60 m at an angle of 45 degrees downward from the horizontal:

- Force = 4.00 * 10^4 N/C
- Distance = 2.60 m
- theta = 45 degrees

Plugging the values into the formula:

Work = (4.00 * 10^4 N/C) * (2.60 m) * cos(45 degrees)

The cosine of 45 degrees is √2/2, so the calculation becomes:

Work = (4.00 * 10^4 N/C) * (2.60 m) * (√2/2)

Calculating the numerical value:

Work ≈ 1.82 * 10^4 J

Therefore, the work done when the charge moves 2.60 m at an angle of 45 degrees downward from the horizontal is approximately 1.82 * 10^4 joules.

To find the work done by the electric force, we can use the formula:

Work = Force × Distance × cos(θ)

where:
- Work is the amount of work done by the force (in joules, J)
- Force is the magnitude of the electric force (in newtons, N)
- Distance is the displacement of the charge (in meters, m)
- θ is the angle between the force and the displacement (in degrees)

Let's calculate the work done in each case:

(a) When the charge moves 0.450 m to the right:
- Force = 4.00 × 10^4 N/C × 28.0 × 10^(-9) C (since Force = Electric Field × Charge)
- Distance = 0.450 m
- θ = 0° (since the force and displacement are in the same direction)

Substituting these values into the formula:
Work = (4.00 × 10^4 N/C × 28.0 × 10^(-9) C) × 0.450 m × cos(0°)

Simplifying and calculating the cosine of 0° (which is 1):
Work = (4.00 × 10^4 N/C × 28.0 × 10^(-9) C) × 0.450 m × 1

You can now calculate the value of Work.

(b) When the charge moves 0.670 m upward:
- Force = 4.00 × 10^4 N/C × 28.0 × 10^(-9) C (same as before)
- Distance = 0.670 m
- θ = 90° (since the force and displacement are perpendicular)

Substituting these values into the formula:
Work = (4.00 × 10^4 N/C × 28.0 × 10^(-9) C) × 0.670 m × cos(90°)

Simplifying and calculating the cosine of 90° (which is 0):
Work = (4.00 × 10^4 N/C × 28.0 × 10^(-9) C) × 0.670 m × 0

You can now calculate the value of Work.

(c) When the charge moves 2.60 m at an angle of 45 degrees downward from the horizontal:
- Force = 4.00 × 10^4 N/C × 28.0 × 10^(-9) C (same as before)
- Distance = 2.60 m
- θ = 45°

Substituting these values into the formula:
Work = (4.00 × 10^4 N/C × 28.0 × 10^(-9) C) × 2.60 m × cos(45°)

Simplifying and calculating the cosine of 45° (which is (√2)/2):
Work = (4.00 × 10^4 N/C × 28.0 × 10^(-9) C) × 2.60 m × (√2)/2

You can now calculate the value of Work.