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

E = 258.84 N / C

Step-by-step 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