Question

Electric Field at a Point

A -70nC charge is distributed uniformly along the x-axis from x = -0.8m to x = 2.6m. Consider a point at y = 1.5m on the y-axis.

a) What is the x-component of the electric field at the point?

b) What is the y-component of the electric field at the point?

c) What is the total magnitude of the electric field at the point?

Thank you

Answer 1

B) Suppose you are now asked to calculate the electric field at point P located a distance b from the side of the uniformly charged rod.

Notice in the following diagram that we must deal with both the horizontal (E_x) and vertical (E_y) components of the electric field at P. Since the vertical components all point in the same direction -- away from the charged rod -- we will start with them.

Using vertical angles and right triangle trigonometry, we can calculate that

Taking the limit as Dx approaches 0, we get that

Unfortunately this leaves us with an expression involving three variables: x, r, and q. Since our differentiable is dx, we need to replace r and q with equivalent expressions involving only x. We can make this happen by noticing the following relationships:

Substituting for r and cos q, we get

where x = 0 is at point O.

Since

After integrating, our final expression for E_y becomes

We will now repeat our process and solve for E_x.

Using vertical angles and right triangle trigonometry, we can calculate that

Again, this leaves us with an expression involving three variables: x, r, and q. Since our differentiable is dx, we need to replace r and q with equivalent expressions involving only x. We can make this happen by noticing the following relationships:

Substituting for r and sin q, we get

Since

After integrating, our final expression for E_x becomes

lamda = q/x = 20.588nC

L -a = 2.6m

b = 1.5m

a = .8m

substituting

E_x = 170.759 N/C

E_y = 165.188N/C

E = SQRT[E_x²+E_y²]

E =237.588N/C