Find the potential difference in a rectangle

In summary, two children are playing in the park. The potential energy of an electrostatic field between two children is greater than the potential energy of two children playing in the park.
  • #1
PhysicsIdiot007
2
0

Homework Statement


[/B]
Figure 20-3, referred to below, is 0.800m wide and 0.400m tall with "A" in the top left corner, "+4 microC" charge in the top right corner, "+2 microC" charge in the bottom left corner, and "B" in the bottom right corner.

Two point charges of magnitude +4.00 μC and +2.00 μC are placed at the opposite corners of a rectangle as shown in Figure 20-3.
(a) What is the potential at point A due to these charges?
(b) What is the potential at point B due to these charges?
(c) What is the potential difference between points A and B?

Homework Equations



U=(kQq)/r where k=8.99E9 Nm^2/C^2

U=qV

V=(kq)/r

The Attempt at a Solution



I honestly do not know where to begin this problem, other wise I wouldn't have posted it. I started by trying to plug values into "U=(kQq)/r" but quickly realized that would only help me if I knew the charge on points A and B.

This is what I had done before I realized I had no idea what I was doing:
U=(8.99E9*4*2)/0.89 (0.89 is the hypotenuse of the rectangle)

I was then going to plug that number into U=qV and solve for V, but I have no q for points A or B so I'm stuck.
 
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  • #2
Here is the Figure
 

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  • #3
Use the formula for potential. This is the quantity you have to calculate.
Calculate potential at A produced by each charge, separately. Then add the two potentials.
 
  • #4
PhysicsIdiot007 said:
potdif.JPG


Homework Equations



U=(kQq)/r where k=8.99E9 Nm^2/C^2
This is the potential energy of a charge q in the field of a a point charge Q. You need the potential, V, the potential energy of a unit positive charge. You wrote correctly that
PhysicsIdiot007 said:
U=qV

V=(kq)/r

The Attempt at a Solution



I honestly do not know where to begin this problem, other wise I wouldn't have posted it. I started by trying to plug values into "U=(kQq)/r" but quickly realized that would only help me if I knew the charge on points A and B.

You need the potential V=kQ/r. that q in the formula for the potential energy is 1 C.

PhysicsIdiot007 said:
This is what I had done before I realized I had no idea what I was doing:
U=(8.99E9*4*2)/0.89 (0.89 is the hypotenuse of the rectangle)

I was then going to plug that number into U=qV and solve for V, but I have no q for points A or B so I'm stuck.
You do not need U. You need V. Recall, that the potential at distance r from a point charge Q is kQ/r.
The potential at A is the sum of the potentials from both the ##2 \mu C## charge and the
##4 \mu C## charge. Substitute the appropriate distances in the formula kQ/r. How far is A from both charges?

ehild
 
  • #5

To calculate the potential difference between points A and B, we first need to determine the potential at each point due to the given charges. To do this, we can use the equation V=(kq)/r, where V is the potential, k is the Coulomb's constant (8.99E9 Nm^2/C^2), q is the charge, and r is the distance between the charge and the point.

(a) To find the potential at point A, we need to calculate the distance between the charge of +4.00 μC and point A. Since the charges are placed at opposite corners of the rectangle, the distance between them is equal to the diagonal of the rectangle, which can be found using the Pythagorean theorem. So, the distance between the charge and point A is √(0.800^2+0.400^2) = 0.894 m.

Now, we can plug in the values into the equation V=(kq)/r:
V = (8.99E9*4.00E-6)/0.894 = 4.01 V

Therefore, the potential at point A due to the given charges is 4.01 V.

(b) To find the potential at point B, we can use the same equation and calculate the distance between the charge of +2.00 μC and point B. Since the charges are placed at opposite corners, the distance between them is also equal to the diagonal of the rectangle, which is 0.894 m.

Plugging in the values, we get:
V = (8.99E9*2.00E-6)/0.894 = 2.00 V

Therefore, the potential at point B due to the given charges is 2.00 V.

(c) The potential difference between points A and B is simply the difference between the potentials at these points. So, we can calculate it by subtracting the potential at point B from the potential at point A:
VAB = VA - VB = 4.01 V - 2.00 V = 2.01 V

Therefore, the potential difference between points A and B is 2.01 V.
 

1. What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential between two points in an electric circuit.

2. How is potential difference calculated?

The potential difference in a rectangle can be calculated by multiplying the electric field strength by the distance between the two points in the rectangle.

3. What is the unit of potential difference?

The unit of potential difference is volts (V) in the SI system.

4. How does potential difference affect the flow of electric current?

Potential difference is directly proportional to the flow of electric current. A higher potential difference will result in a higher current flow, and vice versa.

5. Can potential difference be negative?

Yes, potential difference can be negative. This indicates that the current is flowing in the opposite direction of the electric field.

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