Given an Electric Field find the necessary Work to move a charge Q

In summary, the necessary work to move a charge Q=7\mu C along an incremental path of 1mm distance in the direction of 2\vec{u_{x}}-6\vec{u_{y}}-3\vec{u_{z}} at point P(1,2,3) is -75nJ. When converting to base units, dl should be a unit vector, resulting in a solution of W=-525nJ.
  • #1
kacete
27
0

Homework Statement


Given the Electric field [tex]\vec{E}=z\vec{u_{x}}-3y^{2}\vec{u_{y}}+x\vec{u_{z}} V/m[/tex] find the necessary Work to move a charge [tex]Q=7\mu C[/tex] along an incremental path of [tex]1mm[/tex] distance in the direction of [tex]2\vec{u_{x}}-6\vec{u_{y}}-3\vec{u_{z}}[/tex] which locates in [tex]P(1,2,3)[/tex]

Homework Equations


Work
[tex]W=-Q\int^{f}_{i}\vec{E}.d\vec{l}[/tex]
i - inicial point
f - final point
Solution
[tex]W=-75nJ[/tex]

The Attempt at a Solution


Converting to base units
[tex]Q=7\mu C=7x10^{-6}C[/tex]
[tex]d=1mm=1x10^{-3}m[/tex]
The electric field at the given point
[tex]\vec{E}=3\vec{u_{x}}-12\vec{u_{y}}+1\vec{u_{z}} V/m[/tex]
The equation
[tex]W=-Q\int^{f}_{i}\vec{E}.d\vec{l}=-Q.E\int^{f}_{i}dl.(2\vec{u_{x}}-6\vec{u_{y}}-3\vec{u_{z}})[/tex]
where [tex]\int^{f}_{i}dl=1x10^{-3}[/tex]
My solution
[tex]W=-525nJ[/tex]

Where did I go wrong?
 
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  • #2
You need to make dl a unit vector.
 
  • #3
Can't believe it was that simple. I'm embarrassed. Thank you!
 

1. What is an electric field?

An electric field is a physical quantity that represents the force per unit charge acting on a charged particle. It is created by other charged particles and can be either attractive or repulsive depending on the nature of the charges involved.

2. How is electric field measured?

Electric field is measured in units of Newtons per Coulomb (N/C) in the SI system. It can also be measured in Volts per meter (V/m) or in units of energy per charge (J/C) in the CGS system.

3. What is the relationship between electric field and work?

The work done by an electric field on a charged particle is equal to the product of the electric field strength and the distance over which the particle is moved. This can be expressed mathematically as W = Eqd, where W is the work, E is the electric field strength, q is the charge, and d is the distance.

4. How do you calculate the necessary work to move a charge in an electric field?

The necessary work to move a charge in an electric field can be calculated by multiplying the magnitude of the charge by the potential difference between the two points in the field. This can be expressed as W = QΔV, where W is the work, Q is the charge, and ΔV is the potential difference.

5. Can you have a negative work in an electric field?

Yes, it is possible to have a negative work in an electric field. This occurs when the direction of the force exerted by the electric field is opposite to the direction of motion of the charged particle. In this case, the work done by the electric field is negative, indicating that the particle is losing energy as it moves in the direction opposite to the electric field.

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