Electric Field Calculation for Infinitely Long Charged Line and Rod

In summary: Therefore, the force that exerts on the infinitesimally small part of the rod is:dF = λ(line)*E(y+a)*dywhere λ(line) is the linear charge density of the infinite line, y is the distance from the lower end of the rod and dy is the infinitesimally small part of the rod.Now, we know that the electrical field that the infinite line produces is:E(s) = (λ(line))/(2πε0*s)Therefore, the force is:dF = [(λ(line)*λ(rod)*dy)/(2πε0*(y+a))] The total force acting on the rod is the integration of the force that we have just found. Therefore:F
  • #1
Govyadina
1
0

Homework Statement



An infinitely long, uniformly charged straight line has linear charge density λ1 coul/m. A straight rod of length 'b' lies in the plane of the straight line and perpendicular to it, with its enared end at distance 'a' from the line. The charge density on the rod varies with distance 'y', measured from the lower end, according to λ(on rod) = (λ2*b)/(y+a), where λ2 is a constant. Find the electrical force exerted on the rod by the charge on the infinite straight line, in the λ1, λ2, a, and b, and constants like ε0.


Homework Equations



Gauss's Law.


The Attempt at a Solution



I first treated the problem as if there was only a point P above the infinite line and applied Gauss's Law using a cylinder as my Gaussian surface. My answer was

E = (λ2 * b) / [2pi*ε0*(a+b)^2]

From my understanding, I now have to relate my first answer to a change in distance of the point P along the rod. Any help would be appreciated.

Thank you in advance.
 
Physics news on Phys.org
  • #2
I hope I understood your problem correct!

It's correct to use Gauss law to find the field of the line, but I think that you made a mistake because your answer is independent of [itex]\displaystyle{\lambda _1}[/itex]. Also your answer is not general, because point P is not a random point, it is rod's upper end (its distance from the line is [itex]\displaystyle{a+b}[/itex]).

Now let's see a way to solve the problem.
Firstly use Gauss law to find the field [itex]\displaystyle{E(r)}[/itex] of the line at a random point which distance from the line is a variable [itex]\displaystyle{r}[/itex]. This [itex]\displaystyle{r}[/itex] must be the radius of your Gaussian cylinder.

Then take an infinitesimally small part of the rod, with length [itex]\displaystyle{dy}[/itex], which distance from the lower end of the rod is [itex]\displaystyle{y}[/itex] (and consequently from the line is [itex]\displaystyle{r=y+a}[/itex]). This part is charged with charge [itex]\displaystyle{dq}[/itex], so using [itex]\displaystyle{E(r)}[/itex] we can compute the force [itex]\displaystyle{dF}[/itex] exerted on it. Note that you can also find [itex]\displaystyle{dq}[/itex] in terms of [itex]\displaystyle{y}[/itex] and [itex]\displaystyle{dy}[/itex] using rod's charge density.

Finally you get an equation for [itex]\displaystyle{dF}[/itex] in terms of [itex]\displaystyle{y}[/itex], [itex]\displaystyle{dy}[/itex] and some constants. You can integrate this equation to find the total force on the rod.

I didn't use vectors for the field or the force because I think that the direction is trivial at this problem (everything upwards).
 
  • #3
The charge of the infinitesimally small part of the rod with length dy is λ(rod)*dy. Which is the force that acts on it? since you have found the electrical field that the line produces, then the force that exerts on a charge which lies in distance s from it is: dF = E(s)*dq.
 

1. How are electric fields created?

Electric fields are created by the presence of electric charges. Positive charges create an outward electric field, while negative charges create an inward electric field.

2. What is the formula for calculating electric field strength?

The formula for electric field strength is E = F/q, where E is the electric field strength, F is the force exerted by the field, and q is the magnitude of the charge.

3. How do electric fields affect charged particles?

Electric fields exert a force on charged particles, causing them to accelerate or move in a certain direction. The magnitude and direction of this force depend on the charge of the particle and the strength and direction of the electric field.

4. What is the difference between electric potential and electric field?

Electric potential is a measure of the potential energy per unit charge at a specific point in an electric field, while electric field is a measure of the force per unit charge at a specific point in an electric field.

5. How can electric fields be shielded or blocked?

Electric fields can be shielded or blocked by conducting materials, such as metal, which allow charges to flow freely and neutralize the electric field. Non-conductive materials, such as rubber or plastic, can also block electric fields by preventing the flow of charges.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
784
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
6K
  • Introductory Physics Homework Help
Replies
3
Views
947
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
743
  • Introductory Physics Homework Help
Replies
26
Views
576
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
3K
Back
Top