# Essay Experiment 1

Submitted By kells07
Words: 533
Pages: 3

Experiment: 1

Electric Field & Electric Potential

Electric Field and Electric Potential

Purpose

The objective of this lab is to map the electric equipotential lines and electric field lines for two-dimensional charge configurations.

Theory

The magnitude of electrostatic force between two point charges Q and q is given by Coulomb’s law:
F = k |Qq|/ r2
The magnitude of the electric field is defined as the electrical force per unit charge:
E= (N/C)

The work done by an electric field in moving a unit charge from point A to point B is called the potential difference between these two points:
ΔVAB = VA – VB = WAB /q

Q is the primary source charge r is the distance away from the charge k is 9.0E9 Nm²/C²

Data
1)

2) Yes, the field lines are perpendicular to the equipotential lines,

3)

∆V

∆V12=V1-V2
∆V23=V2-V3
∆V34=V3-V4
∆V45=V4-V5
∆x
0.01m
0.01m
0.01m
0.01m
E=∆V/∆x

154 N/C
152 N/C
166 N/C
189 N/C

4) The potential difference might be due to the experimental error, but also it happens because some energy is lost due to the environment. The points 1A to 1H theoretically are on the same equipotential line.
5)

6)
A) Points 3A to 3G are on the same equipotential line because the potential values are about the same.

B) 1A and 1G are not on the same equipotential line because one is closer to the negative charge than the other one.

C) This happens because the potential values decreases as it gets closer to the negative charge.

7)

8) The essential reason making the difference in shape of the equipotential lines between Fig 3 and Fig 4 is that in figure 3 the positive charge was on the center which made it symmetric; on the other hand, on figure 4 there was no symmetry between the two charges.

Calculations

Average V1= (8.08+8.05+8.23+8.21+8.32)/5 = 8.18V
Average V2= (6.61+6.44+6.71+6.64+6.78)/5 = 6.64V
Average V3= (4.91+5.07+5.09+5.21+5.33)/5 = 5.12V
Average V4=…