Work done by charge in electric field on the xyz plane

Tags:
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