E field at point from graph of V and x?

[asdf12312]

Homework Statement

Homework Equations

E = -dV/dx ?

The Attempt at a Solution

not really sure how to do this problem. I think I understand E field is negative of slope of V/x, but I'm getting it wrong. just looking at it, I see V=-2 at x=7.5. so the E field to right of x= 5 is y2-y1/x2-x1= 2/2.5 = 0.8. to the left of it I guess it's -0.8 (V/m). I tried both answer but its wrong.

 

[collinsmark]

Homework Statement

Homework Equations

E = -dV/dx ?

The Attempt at a Solution

not really sure how to do this problem. I think I understand E field is negative of slope of V/x, but I'm getting it wrong. just looking at it, I see V=-2 at x=7.5. so the E field to right of x= 5 is y2-y1/x2-x1= 2/2.5 = 0.8. to the left of it I guess it's -0.8 (V/m). I tried both answer but its wrong.

Finding the slope is the correct approach. So in that respect you are on the right track.

But the way in which you are going about finding the slope is too inaccurate. There's a better way.

Can you express V(x) in terms of a sine function? In other words, if

V(x) = A sin(kx),

can you find A and k ?

 

[asdf12312]

oh yeah, I didn't think of that. V(x) = 2*sin(0.2*pi*x), so I guess I take the derivative of this to find E field? dV/dx = 2*0.2*pi*cos(0.2*pi*x), and at x=5, E=-dV/dx=1.26?

 

[collinsmark]

oh yeah, I didn't think of that. V(x) = 2*sin(0.2*pi*x), so I guess I take the derivative of this to find E field? dV/dx = 2*0.2*pi*cos(0.2*pi*x), and at x=5, E=-dV/dx=1.26?

That looks correct to me! :)

(Don't forget to include the proper units in the final answer through. [Edit: then again, the problem statement didn't specify the units of V nor x. So nevermind about that, I guess.])

 

[rezeew]

Your understanding of the relationship between the electric field (E) and the potential (V) is correct. The electric field is indeed equal to the negative of the slope of the potential with respect to distance (x). However, there are a few things to consider when solving this problem.

First, make sure you are using the correct units for the potential. The potential is typically measured in volts (V), not meters (m). If the units are not given, make sure to clarify with your instructor.

Second, when calculating the slope, make sure to use the correct points on the graph. In this case, you are given the potential at x=7.5, but you are asked to find the electric field at a point to the left or right of this point. Make sure to use the correct points on the graph to calculate the slope.

Finally, double check your calculation. In this case, the slope should be -0.8 V/m, not 0.8 V/m. This is because the potential is decreasing as you move to the right, so the slope will be negative.

Overall, it seems like you have the right approach, but just need to double check your units and calculations. Keep practicing and you will get the hang of it!