Calculating electric field due to charge distribution

This distance is (a+r-x) since the distance from the left end of the rod to P is (a+x). So the net distance is (a+r-x) and that is what is used in the equation. Therefore, in summary, the solution for part (a) correctly uses the distance (a+r-x) even though q is not explicitly mentioned in the problem.
  • #1
henry3369
194
0

Homework Statement


http://imgur.com/EiCFIDe

Homework Equations


E = kq/r2
λ = dq/dx
Q = λa

The Attempt at a Solution


Here is the book solution:
http://imgur.com/aKSeKo0

I need help with part (a). I understand everything in the solution except for the distance they used for between (r2 in the equation above). It seems they used (a+r-x) and they say that is the distance between a segment at x and the charge q. What does q have to do with the problem at all? Part (a) asks for the electric field due to the charge distribution Q where x > a. So shouldn't the distance be between dx(a segment on the charge distribution) and x?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
You are right, it should not have mentioned q in the solution for (a). But the derivation is OK if you think of r as the distance from the right end of the rod to some arbitrary point P located to the right of the rod.
 

Related to Calculating electric field due to charge distribution

1. How do I calculate the electric field due to a point charge?

To calculate the electric field due to a point charge, you can use the equation E = kQ/r², where E is the electric field, k is the Coulomb's constant (9x10^9 Nm²/C²), Q is the magnitude of the charge, and r is the distance between the charge and the point where you want to calculate the electric field.

2. What is the equation for calculating the electric field due to a continuous charge distribution?

The equation for calculating the electric field due to a continuous charge distribution is E = (1/4πε₀) ∫(ρ/r²) dV, where E is the electric field, ε₀ is the permittivity of free space (8.85x10^-12 C²/Nm²), ρ is the charge density, r is the distance between the charge element and the point where you want to calculate the electric field, and dV is the volume element.

3. Can I use the superposition principle to calculate the electric field due to multiple charges?

Yes, you can use the superposition principle to calculate the electric field due to multiple charges. This principle states that the total electric field at a point is the vector sum of the individual electric fields due to each charge.

4. How do I calculate the electric field at a point if the charge distribution is not symmetric?

If the charge distribution is not symmetric, you can use the principle of superposition to break down the charge distribution into smaller, simpler parts. Then, you can use the equation for calculating electric field due to a continuous charge distribution for each individual part and add them together to get the total electric field at the point of interest.

5. Can I use the electric field equation to calculate the electric potential?

No, the electric field equation E = (1/4πε₀) ∫(ρ/r²) dV is used to calculate the electric field due to a charge distribution. To calculate the electric potential, you can use the equation V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the magnitude of the charge, and r is the distance between the charge and the point where you want to calculate the potential.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
805
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
584
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
23
Views
624
  • Introductory Physics Homework Help
Replies
2
Views
441
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
751
  • Introductory Physics Homework Help
Replies
3
Views
838
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top