Constant Electric Field and Potential

In summary, the conversation is about solving for the potential at Point 2 in an electric field given the coordinates and magnitude of the field. The solution involves using the formula Vf-Vi = -Integral from i to f (E*ds) and plugging in the given values to solve for Vf. The initial potential at Point 1 is 1200 V and the distance between the two points is found to be the square root of 106 meters. However, the final answer of 1148 V was incorrect due to a mistake in plugging in the numbers.
  • #1
ganondorf29
54
0

Homework Statement



The figure below shows two points in an electric field. Point 1 is at (X1,Y1) = (3,4), and point 2 is at (X2,Y2) = (12,9). (The coordinates are given in meters.) The electric field is constant with a magnitude of 59.3 V/m, and is directed parallel to the +X-axis. The potential at point 1 is 1200.0 V.

prob02a_UEcoord34129.gif




Homework Equations



Vf-Vi = -Integral from i to f (E*ds)



The Attempt at a Solution



I set Vi=1200V
E = 59.3 N/C
and tried to solve for Vf

I found the distance by making a triangle connecting the two points to be the sqrt(106) and the angle formed to be 29deg. I took the cos(29) and brought that out of the integral(Im not sure if that's right, but I saw my book take the cosine of the angle so I did too).

Vf-1200 = -59.3*cos(29)*integral(from i to f) of (ds)

I solved for Vf and got 1148 V, but that's wrong. Any suggestions
 
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  • #2
ganondorf29 said:
Vf-1200 = -59.3*cos(29)*integral(from i to f) of (ds)

I solved for Vf and got 1148 V, but that's wrong. Any suggestions

Looks like you just plugged the numbers in wrong. I assume that you recognized [tex]\int_i^fds=\sqrt{106}[/tex]?
 
  • #3
?

I would like to offer some feedback on your solution attempt. First, your approach seems generally correct, but there are a few things to consider.

1. When using the equation Vf-Vi = -Integral from i to f (E*ds), it is important to make sure all units are consistent. In this case, the electric field is given in V/m, but the distance is given in meters. The units should cancel out in the end, but it's always good practice to double check.

2. Your approach of using trigonometry to find the distance and angle is correct. However, it may be easier to use Pythagorean theorem to find the distance (sqrt(9^2 + 5^2) = sqrt(106)) instead of making a triangle. Also, the angle you found (29 degrees) is not the correct angle. Remember that the electric field is parallel to the +X-axis, so the angle between the electric field and the distance vector should be 0 degrees.

3. The integral should be taken from point 1 to point 2, not from i to f. This means that the limits of integration should be the coordinates of point 1 and point 2, not just i and f.

Taking these into consideration, the correct solution would be:

Vf - 1200 = -59.3 * cos(0) * integral(from 3 to 12) of ds

Vf = 1200 - 59.3 * (12-3) = 1147.1 V

I hope this helps and good luck with your homework!
 

Related to Constant Electric Field and Potential

1. What is a constant electric field?

A constant electric field is a region of space where the electric field strength remains the same at all points. This means that the force experienced by a charged particle placed in the field will be the same regardless of its position.

2. How is a constant electric field created?

A constant electric field can be created by placing two oppositely charged parallel plates close together. The electric potential difference between the plates creates a uniform electric field between them.

3. What is the relationship between electric field and potential in a constant electric field?

In a constant electric field, the potential is directly proportional to the electric field strength. This means that the potential decreases as the distance from the source of the field increases.

4. What is the significance of a constant electric field and potential in practical applications?

Constant electric fields and potentials are important in many practical applications, such as in electronic devices and power transmission. They allow for the controlled movement of charged particles and the efficient transfer of energy.

5. How is the strength of a constant electric field measured?

The strength of a constant electric field is measured in units of volts per meter (V/m). This represents the electric field strength experienced by a charge of 1 volt at a distance of 1 meter from the source of the field.

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