Electric Potential Question

[dect117]

1. Homework Statement :

A graph of the x component of the electric field as a function of x in a region of space is shown in the figure [attached] below. The scale of the vertical axis is set by E_xs = 47.5 N/C. The y and z components of the electric field are zero in this region. Suppose that the electric potential at the origin is 14 V.

(a) What is the electric potential at x = 2.0 m
(b) What is the greatest positive value of the electric potential for points on the x axis for which
0 ≤ x ≤ 6.0 m?
(c) For what value of x is the electric potential zero?
​

Homework Equations

:[/B]
$$V_f-V_i=-\int_{i}^{f} \vec E \cdot d \vec s$$

The Attempt at a Solution

:[/B]

For part (a), I began by substituting in the appropriate values.
$$V_f-14=-\int_{0}^{2} E~d x$$
After integrating, I got
$$V_f-14=-E\left( 2 \right)$$
Finally, I substituted -47.5 in for E and solved for V_f.
$$V_f=47.5\left( 2 \right)+14=109$$
The answer I got was incorrect, though, and I can't see where I went wrong. I haven't really attempted parts (b) or (c) yet.

 

[gneill]

I think you need to check how you performed your integration. The value of E is not constant over the distance.

Hint: Think of the integral as the area "under" the curve. In the case of part (a), the area of interest stretches from x = 0 to x = 2 m.

 

[dect117]

I think you need to check how you performed your integration. The value of E is not constant over the distance.

Hint: Think of the integral as the area "under" the curve. In the case of part (a), the area of interest stretches from x = 0 to x = 2 m.

View attachment 222429

Oh crap, I completely ignored that the electric field isn't constant. So if the integral of dx is the area of a triangle, would the height just be 1?

 

[gneill]

Oh crap, I completely ignored that the electric field isn't constant. So if the integral of dx is the area of a triangle, would the height just be 1?

The "height" is determined by the scale of the vertical axis, and where on the curve your triangle vertex lies.

 

[dect117]

The "height" is determined by the scale of the vertical axis, and where on the curve your triangle vertex lies.

So h = -47.5? If so, I get $$V_f-14=-\frac 1 2 E b h=-\frac 1 2 \left( -47.5 \right)^2 \left( 2 \right)$$ which doesn't yield the correct answer. Isn't the value of E also -47.5?

 

[dect117]

If h = 1, then I get $$V_f-14=-\frac 1 2 \left( -47.5 \right) \left( 2 \right)=47.5$$ $$V_f=47.5+14=61.5$$ which WebAssign says is the correct answer to part (a). I am very confused. lol

 

[gneill]

If h = 1, then I get $$V_f-14=-\frac 1 2 \left( -47.5 \right) \left( 2 \right)=47.5$$ $$V_f=47.5+14=61.5$$ which WebAssign says is the correct answer to part (a). I am very confused. lol

I don't understand your use of "h". In your work you use -47.5 N/C for the "height" of the triangle, not 1 N/C.

 

[dect117]

I don't understand your use of "h". In your work you use -47.5 N/C for the "height" of the triangle, not 1 N/C.

E = -47.5, no? If so, then -47.5 will be squared if h is also equal to -47.5.

 

[gneill]

E = -47.5, no? If so, then -47.5 will be squared if h is also equal to -47.5.

Um, no. If the vertex is at E = -47.5 N/C, then that is the "height" of the triangle. h is not a separate variable here, it simply represents the height of the triangle under consideration. In this case, for part (a) of the problem, the height of the triangle happens to be -47.5 N/C.