Question

A charge per unit length λ = +6.00 μC/m is uniformly distributed along the positive y-axis from y = 0 to y = +a = +0.400 m. A charge per unit length λ = -6.00 μC/m, is uniformly distributed along the negative y-axis from y = 0 to y = –a = -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

consider a small charge element of length 'dy' at a distance of 'y' from the origin

small charge on small length is given as

dq = dy

small electric field by this small charge element at point A is given as

dE = k dq /r²

dE = k dy /(x² + y²)

In triangle AOB

Sin = y/r = y/sqrt(x² + y²)

Electric field components in X-direction cancel out and we get net electric field in Y-direction as

dE' = 2 dE Sin = 2 (k dy /(x² + y²)) (y/sqrt(x² + y²))

dE' = 2 kydy / (x² + y²)^3/2

Net electric field is given as

E = - dE'

E = (-2 k ) ydy / (x² + y²)^3/2

E = (-2 k )

E = (- 2 x 9 x 10⁹ x 6 x 10^-6)

x = 0.431

E = (-108000) ((1/sqrt((0.271)² + 0.4²)) - (1/sqrt((0.271)² + 0²))

E = - 1.75 x 10⁵ N/C

So magnitude = 1.75 x 10⁵ N/C