Question

A charge per unit length λ = -8.00 μC/m, is uniformly distributed along the negative y-axis from y= 0 to y = âa = -0.400 m. What is the magnitude of the electric field at a point on the x-axis a distance x = 0.271 m from the origin?

Answer 1

Answer:

 E = 258.84 N / C

Explanation:

The electric field of a continuous charge distribution is

            E = k ∫ dQ / r²

Where k is the Coulomb constant that you are worth 8.99 10⁹ N m²/C², Q is the load and r the distance between the charges and the point of interest

Let's look for the distance with Pythagoras' theorem

            r² = x² + y²

For the load we use the concept of linear load density

            λ = Q / y = dQ / dy

           dQ =  λ dy

We substitute in the first equation

           E = k ∫ λ dy / (x² + y²)

This integral is immediate (Tabulated)

          ∫ dy / (a² + y²) = 1 /a  tan⁻¹ (y / a)

Substituting in our equation results in

            E = k Lam /x  tan⁻¹ (y / x)

Let's evaluate betwe¹en the lower limit y = 0 and the upper limit y = -0.400 m

            E = k  λ / x [tan⁻¹ (-0.4 / a) –tan⁻¹ 0]

           E = k  λ/x tan⁻¹ (-0.4 / a)

Let's calculate

          E = 8.99 10⁹ (-8.00) /0.271 tan⁻¹ (-0.4 / 0.271)

The angle of the tan-1 must be in radians

           E = 265.387 0.9753

          E = 258.84 N / C