Calculating Potential Difference in a Uniform Electric Field | Example Problem

In summary, the potential difference between the points (0cm, -5cm) and (1cm, 4cm) in a uniform electric field with a magnitude of 20000 V/m and direction of 53.13 degrees is -4876.5 V. This is calculated using the dot product of the electric field vector (20000i - 50000j) V/m and the displacement vector (1cm, 9cm), resulting in a negative potential difference of 4876.5 V.
  • #1
KillerZ
116
0
I am wondering if I did this right.

Homework Statement



What is the potential difference between the points (x_i, y_i) = (0cm, -5cm) and (x_f, y_f) = (1cm, 4cm) in a uniform electric field equal to E = (20000i - 50000j) V/m ?

Homework Equations



[tex]\Delta V = V(s_{f})-V(s_{i}) = -\int^{s_{f}}_{s_{i}}E_{s}ds[/tex]

E is uniform therefore:

[tex]\Delta V = - E_{s}\Delta s[/tex]

[tex]\Delta s = \sqrt{(9cm)^{2}+(1cm)^{2}}[/tex]

[tex]= \frac{\sqrt{82}}{100} m[/tex]

[tex]E = \sqrt{(20000V/m)^{2}+(-50000V/m)^{2}}[/tex]

[tex]= \sqrt{2.9*10^{9}} V/m[/tex]

The Attempt at a Solution



[tex]\Delta V = - E_{s}\Delta s[/tex]

[tex]= -(\sqrt{2.9*10^{9}} V/m)(\frac{\sqrt{82}}{100} m)[/tex]

[tex]= -4876.5 V[/tex]
 
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  • #2
Don't you have to take the angle into account? Only a component of E is in the direction of the distance.
 
  • #3
I think you take your

ΔV = E*Δs a little differently. Namely as the dot product of the E vector and the s vector, such that

ΔV = Ex*Δx i + Ey*Δy j
 

FAQ: Calculating Potential Difference in a Uniform Electric Field | Example Problem

What is potential difference in a uniform electric field?

Potential difference, also known as voltage, is the difference in electric potential energy per unit charge between two points in a uniform electric field. It is measured in volts (V).

How is potential difference calculated in a uniform electric field?

Potential difference can be calculated by multiplying the strength of the electric field (in volts per meter) by the distance between the two points (in meters). The formula is V = Ed, where V is the potential difference, E is the electric field strength, and d is the distance between the points.

Can potential difference be negative in a uniform electric field?

Yes, potential difference can be negative in a uniform electric field. This occurs when the direction of the electric field is opposite to the direction of the movement of a positive charge. In this case, the potential energy of the charge decreases as it moves from a point of high potential to a point of low potential.

What unit is used to measure potential difference?

Potential difference is measured in volts (V), which is equivalent to joules per coulomb. In some cases, it may also be measured in electron-volts (eV), especially in the field of particle physics.

How is potential difference related to electric potential energy?

Potential difference is directly proportional to electric potential energy. This means that as the potential difference increases, the potential energy of a charge also increases. The formula for electric potential energy is U = qV, where U is the potential energy, q is the charge, and V is the potential difference.

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