Line of Charge

1. Line of Charge Continuation

2. Class Objectives • Find the electric field strength due to a continuous charge distribution. • Develop problem solving techniques for continuous charge distributions.

3. Student Objectives • Be able to identify the type charge distribution for a given problem. • Be able to use diagrams to describe the problem. • Be able to identify and use symmetry in problems. • Be able to set out questions clearly and to develop problem solving strategies.

4. Line of Charge • Consider a rod with total charge +Q. What is the total electric field a distance z along an axis through its centre?

5. Line of Charge • Consider a rod with total charge +Q. What is the total electric field a distance z along an axis through its centre? Z -L L + + + + + + + + + + + + + + + + + + + 2L

6. Z r x -L L + + + + + + + + + + + + + + + + + + + Line of Charge • Consider a rod with total charge +Q. What is the total electric field a distance z along an axis through its centre? dx 2L

7. Line of Charge • Solution • Understand the geometry. • Determine the easiest way to span the geometry. • Evaluate the contribution due to an infinitesimal charge element. • Exploit symmetry if possible. • Set up and solve integral.

8. Line of Charge • We break up the line into two regions –L to 0 and 0 to +L.

9. Line of Charge • We break up the line into two regions –L to 0 and 0 to +L. • The easiest way to span the line is to integrate over x. • Because of symmetry we could integrate from x=0 to x=L and multiply the result by 2. • However we will integrate from –L to L.

10. Line of Charge • The line of charge is uniformly spread over the rod so that, λ=Q/L (charge per unit length).

11. Line of Charge • The line of charge is uniformly spread over the rod so that, λ=Q/L (charge per unit length). • Therefore a charge element dq of width dx, has charge: dq = λdx.

12. Z r x -L L + + + + + + + + + + + + + + + + + + + Line of Charge • The line of charge is uniformly spread over the rod so that, λ=Q/L (charge per unit length). • Therefore a charge element dq of width dx, has charge: dq = λdx.

13. Z r x -L L + + + + + + + + + + + + + + + + + + + Line of Charge • The line of charge is uniformly spread over the rod so that, λ=Q/L (charge per unit length). • Therefore a charge element dq of width dx, has charge: dq = λdx. dq width dx

14. Line of Charge • Recall: The field at a point due to a point charge is:

15. Z r x -L L + + + + + + + + + + + Line of Charge • The field set up at the point due to a small line element dq is dE. dE

16. Z r x -L L + + + + + + + + + + + Line of Charge • The field set up at the point due to a small line element dq is dE. • From symmetry the horizontal components from the fields produced by elements on the opposite sides of the centre cancel. dE

17. Line of Charge • Method 1: Integrating from –L to L (ignore symmetry).

18. Line of Charge • Therefore the net electric field is:

19. Line of Charge • Therefore the net electric field is: • Summing over the line from x=-L to x=L gives the total electric field due to the line.

20. Z r x -L L + + + + + + + + + + + Line of Charge dE • However we are not yet ready to integrate since we have two variables in our integral.

21. Z r x -L L + + + + + + + + + + + Line of Charge dE • However we are not yet ready to integrate since we have two variables in our integral x,θ.

22. Z r x -L L + + + + + + + + + + + Line of Charge dE • However we are not yet ready to integrate since we have two variables in our integral x,θ. • So that our integral is in terms of x we substitute for cosθ. • From the diagram:

23. Z r x -L L + + + + + + + + + + + Line of Charge dE • However we are not yet ready to integrate since we have two variables in our integral x,θ. • So that our integral is in terms of x we substitute for cosθ. • From the diagram:

24. Line of Charge • Therefore we can write that:

25. Line of Charge • Therefore we can write that: • Where z is a constant.

26. Line of Charge • Therefore we can write that: • Where z is a constant.

27. Line of Charge • Therefore we can write that: • Where z is a constant.

28. Line of Charge • Therefore we can write that: • Where z is a constant. • We now can integrate.

29. Line of Charge • We integrate by substitution.

30. Line of Charge • We integrate by substitution. • Using the substitution we get:

31. Line of Charge • We integrate by substitution. • Using the substitution we get:

32. Line of Charge • We integrate by substitution. • Using the substitution we get:

33. Line of Charge • We integrate by substitution. • Using the substitution we get:

34. Line of Charge • Instead of changing limits we note that:

35. Line of Charge • Instead of changing limits we note that: • We can therefore write,

36. Line of Charge • Instead of changing limits we note that: • We can therefore write,

37. Line of Charge • Considering the case z>>L,

38. Line of Charge • Considering the case z>>L, • Since , then

39. Line of Charge • Considering the case z>>L, • Since , then • The field due to a point charge.

40. Line of Charge • Method 2: Use symmetry and integrate from 0 to L, multiplying the result by 2.

41. Line of Charge • Therefore the net electric field is:

42. Line of Charge • Therefore the net electric field is: • Summing over the line from x=0 to x=L gives the total electric field due to the line.

43. Z r x -L L + + + + + + + + + + + Line of Charge dE • However we are not yet ready to integrate since we have two variables in our integral.

44. Z r x -L L + + + + + + + + + + + Line of Charge dE • However we are not yet ready to integrate since we have two variables in our integral x,θ.

45. Z r x -L L + + + + + + + + + + + Line of Charge dE • However we are not yet ready to integrate since we have two variables in our integral x,θ. • So that our integral is in terms of x we substitute for cosθ.

46. Z r x -L L + + + + + + + + + + + Line of Charge dE • However we are not yet ready to integrate since we have two variables in our integral x,θ. • So that our integral is in terms of x we substitute for cosθ. • From the diagram:

47. Z r x -L L + + + + + + + + + + + Line of Charge dE • However we are not yet ready to integrate since we have two variables in our integral x,θ. • So that our integral is in terms of x we substitute for cosθ. • From the diagram:

48. Line of Charge • Therefore we can write that:

49. Line of Charge • Therefore we can write that: • Where z is a constant.

50. Line of Charge • Therefore we can write that: • Where z is a constant.