Calculating electric potential with a uniform elect field

[Nick Jarvis]

Homework Statement

In a certain region of space, there is a uniform electric field of magnitude
25.0 V m^-1 directed at 30.0 degrees to the direction of the positive x-axis and at 60.0 degrees to the direction of the positive y-axis. (There is no z component of the electric field.) The electric potential at the origin is +150 V. What is the electric potential at the point where x = 5:00 m and y = 12:0 m?[/B]

Homework Equations

I know this is an easy question, but I cannot get my head around it. In the first instance I used:

25.0Vm^-1 = potential difference/distance
Using Pythag, I get the distance to be sqr(5^2 + 12^2) = 13.0m

I know I need to resolve the electric field horizontally and vertically (otherwise I wouldn't have been given the angles). This is exam preparation so I have the answer, but it doesn't give the workings out. If it was all along the horizontal, my answer would be -175V. The answer is NOT this for obvious reasons.

The Attempt at a Solution

[/B]
I tried resolving the electric field along the x and y. 25cos30 = 21.7 and 25sin30 = 12.5

Then using 21.7V^m-1 x 5.0m = 108.5V along the x direction and
12.5V^m-1 x 12.0m = 150.0V, then I tried to using pythag again but my answer does not match with the correct answer. I know I am making it so much more complicated than it needs to be.

Hopefully I have provided the correct info.

Thanks

 

[BvU]

then I tried to using pythag again

Pyth is nice for adding vectors. Here you want to add scalars (the potentials). With ##\Delta V = -\int \vec E \cdot d\vec r \ \ \ ## I get -108 Volt...

 

[Nick Jarvis]

Thanks BvU. Have I done anything correctly? You're spot on with the result, I just cannot work out how to get there. Cheers

 

[Chestermiller]

You have $$dV=-E_xdx-E_ydy$$This is independent of path. So, first you can integrate with respect to x at constant y, and then you can integrate with respect to y at constant x (or vice versa).

 

[BvU]

Have I done anything correctly?

Yes you have: the two numbers (scalar values for the potential differences) are just fine. All you have to do is give them the right sign (a minus sign for both) add them up ! As numbers, not as vectors. With the 150 V at the origin the sum total is -108

 

[Nick Jarvis]

Thanks v much. I cannot find this in my notes at all. Assuming the -108V is due to 3sf? Rather than my 108.5V?

So 150V at the origin. I then resolve in the x and y direction. Cannot work out though why it is 150V - 150V -108V, when we're looking at the +x/y directions? Like I said, this should be such an easy question, but I am really struggling with it.

Many thanks

 

[Chestermiller]

Thanks v much. I cannot find this in my notes at all. Assuming the -108V is due to 3sf? Rather than my 108.5V?

So 150V at the origin. I then resolve in the x and y direction. Cannot work out though why it is 150V - 150V -108V, when we're looking at the +x/y directions? Like I said, this should be such an easy question, but I am really struggling with it.

Many thanks

V=150-21.7x-12.5y

 

[BvU]

I cannot find this in my notes at all

There must be something relating ##V## and ##\vec E##, though ?
Do you recognize/understand my ##
\Delta V = -\int \vec E \cdot d\vec r \ \ \ ## and Chet's ##\ \ dV=-E_x\,dx-E_y\,dy \ \ ## ?

Once you do it's easy to find $$ \begin{align} V(5,12) & = V(0,0) + \Delta V \\ & = V(0,0) -\int \vec E \cdot d\vec r \\ & =V(0,0) - \int_0^5 \int_0^{12} \left ( E_x \, dx + E_y\,dy \right ) \\ & = V(0,0) -\int_0^5 E_x \, dx - \int_0^{12} E_y\,dy \end{align} $$

 

[Nick Jarvis]

Thank-you both. For some reason I was making it way more complicated than it needed to be. Going to work through a few more examples now.

Again, a huge thanks :)

 

[Chestermiller]

Alternatively,

The vector ##\mathbf{x}## is given by $$\mathbf(x)=13 (\sin{(67.38)}\mathbf{i}+\cos{(67.38)}\mathbf{j})$$ and the electric field ##\mathbf{E}## is given by
$$\mathbf{E}=25(\sin{(30)}\mathbf{i}+\cos{(30)}\mathbf{j})$$ So the dot product of ##\mathbf{E}## and ##\mathbf{x}## is given by:$$\mathbf{E}\centerdot{x}=325(\sin{(67.38)}\sin{(30)}+\cos{(67.38)}\cos{(30)})=325 \cos {(67.38-30)}=325\cos{(37.38)}$$This is just the product of the magnitudes of the two vectors times the cosine of the angle between them.