Electrostatic Potential in Electric Fields

In summary, the electrostatic potential at the center of the square is zero, and the electric field strength at the center of the square is also zero.
  • #1
krosenberg
6
0

Homework Statement


1. Four charged particles are held fixed at the corners of a square of side s. All the charges have the same magnitude Q, but two are positive and two are negative. In Arrangement 1, shown below, charges of the same sign are at opposite corners. Express your answers to parts a and b in terms of the given quantities and fundamental constants.
a. For Arrangement 1, determine the following.
i. The electrostatic potential at the center of the square
ii. The magnitude of the electric field at the center of the square
2160792296_e2f3df4057_o.gif


2. The bottom two charged particles are now switched to form Arrangement 2, shown below, in which the positively charged particles are on the left and the negatively charged particles are on the right.
b. For Arrangement 2, determine the following.
i. The electrostatic potential at the center of the square
ii. The magnitude of the electric field at the center of the square
2160017175_a892b8f135_o.gif



Homework Equations


V = PE / q


The Attempt at a Solution


1) I am thinking the the potential at the center, as well as the electric field strength, is zero, since the positives and negatives cancel each other out.
 
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  • #2
Arrangement 1: Correct!

Electric field strength is a vector or a scalar?
 
  • #3
vector; I am not sure how to express my answer for arrangement 2 with the variables (i get frightened without numbers!)
Thanks
 
  • #4
Plot 4 vectors, one for every electric field strength. The magnitude of each field is

[tex] E=k\,\frac{|Q|}{r^2}[/tex]

can you calculate the distance r, from the enter to each charge?
 
  • #5
using the Pythagorean theorem, would r = sqrt(1/2) x s ? or am i going in the completely wrong direction? its really hard to focus on this stuff during christmas break :)
 
  • #6
Actually [tex]r=\sqrt{\frac{s}{2}}[/tex] since it's side is s.

Did you plot the vectors at the center?
 
  • #7
yes, would it be like this? (attached .gif)
Picture 1.gif
 
  • #8
If there are 4 vectors, two pointing at the one [tex]-Q[/tex] charge and two more pointing at the other [tex]-Q[/tex] charge it is correct!

I can't see the attachment yet! :smile:
 

Related to Electrostatic Potential in Electric Fields

1. What is electrostatic potential?

Electrostatic potential is the amount of work required to bring a unit positive charge from infinity to a specific point in an electric field. It is a measure of the potential energy of a charge in an electric field.

2. How is electrostatic potential different from electric potential energy?

Electric potential energy is the potential energy stored in a system of charges due to their positions in an electric field. Electrostatic potential, on the other hand, is the potential energy per unit charge at a specific point in an electric field.

3. How is electrostatic potential calculated?

Electrostatic potential is calculated using the formula V = kQ/r, where V is the electrostatic potential, k is the Coulomb's constant, Q is the magnitude of the charge, and r is the distance from the charge.

4. What is the unit of electrostatic potential?

The unit of electrostatic potential is volts (V), which is equivalent to joules per coulomb (J/C).

5. What is the relationship between electrostatic potential and electric field?

The electric field is the negative gradient of the electrostatic potential (E = - ∇V). This means that the electric field points in the direction of decreasing electrostatic potential. In other words, the electric field shows the direction in which a positive charge would move if placed in the given electric field.

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