Work done by charge in electric field on the xyz plane

[PhysicsQuest1]

Homework Statement

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

Homework Equations

W = qE.d

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.

 

[Geofleur]

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

 

[PhysicsQuest1]

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

Could you tell me if this is right?

E = 270 Cos 58^oi + 270 Cos 32^ok
= 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?

 

[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.

 

[PhysicsQuest1]

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.

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)

 

[Geofleur]

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

 

[PhysicsQuest1]

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

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

 

[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.

 

[PhysicsQuest1]

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.

Ok thanks for your help!