Question

Field at End of Line of Charge A charged rod of length L = 5.60 m lies centered on the x axis as shown. The rod has a linear charge density which varies according to λ = ax where a =-583 C/m2 -L/2 +L/2 What is the total charge on the rod? 0.0000 C 4pts You are correct. Your receipt no. is 154-6850.) PreviousTries What is the x component of the electric field at a point on the x axis a distance of D = 7.80 m from the end of the rod? 4pts Submit Answer Tries 0/10

Answer 1

To calculate total charge on the rod and the electric field at P, we have to consider an elemental length on the rod at a distance 'x' from the left end of the rod (i.e. from origin) .

To calculate total charge on the rod and the electric field at P, we have to consider an elemental length dx on the rod at a distance 'x' from the left end of the rod (i.e. from origin) Given, the linear charge density of rod is lambda = ax (a = -58.3 mu C/m^2 = -58.3 times 10^- 6 C/m^2) So, charge on the elemental length dx will be dq = lambda dx = ax dx So, total charge on the rod will be Q = integral^L_0 dq = integral^L_0 ax dx = [ax^2/2]^L_0 = aL^2/2 Putting the values of a and L, we get Q = -58.3 times 10^- 6 times (5.60)^2/2 = -9.14 times 10^- 8 C Hence, the total charge on rod is - 9.14 times 10^- 8 C Electric Field at P: Electric field due to a point charge 'q' at a distance 'r' can be calculated as E = 1/4 pi elementof_0

Given, the linear charge density of rod is

   ()

So, charge on the elemental length dx will be

  

So, total charge on the rod will be

Putting the values of a and L , we get

Hence, the total charge on rod is -9.14 x 10^-8 C .

Electric Field at P :

Electric field due to a point charge 'q' at a distance 'r' can be calculated as

Where,

  

But for this case we can not use this equation directly because the charge is not fixed at a point (or it is varying with x).

So, electric field at P due to the elemental charge () will be

Where, r = L + D - x (as you can see in the figure)

So,   

  

  

  

  

Putting all the values , we get

(only in x - direction )

Te negative sign is indicating that the electric field is in negative x (i.e. -x ) direction.

NOTE :

The integration :

If

  

Let, a - x = t

or, x = a - t

or, dx = - dt

then,

  

or,   

For any doubt please comment and please give an up vote. Thank you.