# Homework Help: Work done by charge in electric field on the xyz plane

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1. Sep 30, 2015

### PhysicsQuest1

1. The problem statement, all variables and given/known data

There is a uniform electric field, E = 270 N/C, parallel to the xz plane, making an angle of 32o with the positive z axis and an angle of 58o with the positive x axis. A particle with charge q= 0.475 C is moved from the point (xi = 2.0 cm, yi = 0, zi = 0) to the point (xf = 8.5 cm, yf = 6.0 cm, zf = -5.5 cm). How much work is done by the electric field.

2. Relevant equations

W = qE.d

3. The attempt at a solution

The only change in distance I need is parallel to the electric field in the x-z plane.

I have no idea how to find the distance. This is what I tried.

d = (0.055m)2+(0.065m)2 = 0.085 m

Then I believe it's a matter of substituting this into the above equation with the angle which I also don't know how to find. Some guidance would be appreciated thanks.

2. Sep 30, 2015

### Geofleur

How about expressing the displacement and the field vectors in terms of their components? Remember, you can also write $\mathbf{A}\cdot\mathbf{B} = A_xB_x + A_yB_y + A_zB_z$.

3. Sep 30, 2015

### PhysicsQuest1

Could you tell me if this is right?

E = 270 Cos 58oi + 270 Cos 32ok
= 143.08 i + 0 j + 228.97 k

If that's my electric field I basically subtract the distance points to get (d) and put it in W = qE.d to get W. Then I can square the values, add them and square root it to get the magnitude of W?

4. Sep 30, 2015

### Geofleur

Your components for $\mathbf{E}$ look good, but note that $W$ is a scalar, not a vector. $W = \Delta \mathbf{r} \cdot q\mathbf{E}$, and the dot product always yields a scalar.

5. Sep 30, 2015

### PhysicsQuest1

So that means

Since my $\Delta \mathbf{r}$ = (0.065 i + 0.06 j -0.055 k)

Therefore W = | (0.065 i + 0.06 j -0.055 k) $\cdot$ 0.475 (143.08 i + 0 j + 228.97 k)|
= |4.4176 - 5.9818|
= 1.56 J

Am I correct here? (btw Thanks for your quick responses)

6. Sep 30, 2015

### Geofleur

Why take the absolute value? Nothing says that the work has to be positive.

7. Sep 30, 2015

### PhysicsQuest1

Oh yeah good point. So W = -1.56 J. I guess I did it right?

8. Sep 30, 2015

### Geofleur

As a check, you could use the fact that $\mathbf{E}\cdot\Delta\mathbf{r} = E\Delta r\cos\theta$ to get the angle between $\mathbf{E}$ and $\Delta \mathbf{r}$. If the angle is more than 90 degrees, you should expect the work to be negative for a positive charge, like in this case. If you throw a rock upward, then while the rock is traveling upward, gravity is performing negative work on it. Same goes for a positive charge moving against the electric field.

9. Sep 30, 2015