Finding electric field via potential

In summary, a plastic rod of length 1m has a non-uniform charge density λ=cx where c is a positive constant of 2x10^-6 [some unit]. To find the electric field at a point 1m to the left from the left end of the rod, the electric potential must first be calculated using the equation V = k integral(0 to L) dq/r. Then, using the equation E = -dV/dx, the electric field can be determined. The direction of the field is towards the left, and the correct value for the electric field is -1.45 x 10^4 N/C.
  • #1
DannyPhysika
30
0

Homework Statement



The plastic rod of the length L=1 m has a non-uniform charge density λ=cx
where positive constant c =2x10^-6 [some unit]. What unit c has to have? Find the electric
potential at the point on the x-axis 1 m to the left from the left end of the rod. Find the
electric field at that point via potential. What is the direction of the field

Homework Equations



V = k integral(0 to L) dq/r

E = -dV/dx

The Attempt at a Solution



Found everything except electric field. I've tried and gotten so confused as to how to implement it. I got this so far:

dV = kλdx/(d+x)

dV/dx = kλ/(d+x)

E = -k(c)(1m)/(2m) = -8.99 x 10^3 N/C

I know it's wrong but I'm so lost as to where to fix it. Thanks
 
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  • #2
I would like to know if either:

-3.47 x 10^3 N/C

or

-1.45 x 10^4 N/C

Are correct. Thank you :)
 

1. How do you find the electric field from a given potential?

The electric field can be found by taking the negative gradient of the potential. This means finding the derivative of the potential with respect to each coordinate (x, y, z) and then multiplying by -1.

2. What is the formula for calculating electric field from potential?

The formula for calculating electric field from potential is E = -∇V, where E is the electric field, V is the potential, and ∇ is the gradient operator.

3. Can you explain the concept of electric potential and how it relates to electric field?

Electric potential is a measure of the potential energy per unit charge at a specific point in an electric field. It is related to electric field through the equation V = -∫E · dr, where V is the potential, E is the electric field, and the integral is taken along a path from a reference point to the specific point.

4. How does the distance between charges affect the electric field and potential?

The electric field and potential are inversely proportional to the distance between charges. This means that as the distance between charges increases, the electric field and potential decrease.

5. What are the units of electric field and potential?

The units of electric field are volts per meter (V/m) and the units of potential are volts (V). These units are derived from the fundamental units of charge (C), distance (m), and energy (J).

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