Can you please help with these 2 questions ============== Second question A line

Question

Can you please help with these 2 questions

==============

Second question

A line of charge with uniform density of 34.0 nC/m lies along the line y = -15.0 cm, between the points with coordinates x = 0 and x = 48.0 cm. Find the electric field it creates at the origin. Magnitude N/C Direction degree (counterclockwise from the +x axis) A line of charge starts at x = +X0 and extends to positive infinity. The linear charge density is lambda = lambda 0x0/x, where lambda 0 is a constant. Determine the electric field at the origin. (Use any variable or symbol stated above along with the following as necessary: ke.)

Explanation / Answer

1) Ex = integral of k*lambda*Y*dx/(x^2+Y^2)^(3/2) = k*lambda*Y*x/Y^2(x^2+Y^2)^0.5

==> Ex = (9*10^9*34*10^-9/0.15)*((0.48/(0.48^2+0.15^2)^0.5) - 0) = 1947.14 N/C

Ey = integral of k*lambda*x*dx/(x^2+Y^2)^(3/2) = -k*lambda/(sqrt(x^2+y^2)

==> Ey = -(9*10^9*34*10^-9/0.15)*(1/(sqrt(0.48^2+0.15^2)-1/0.15) = 330.96 N/C

E = sqrt(1947.14^2+330.96^2) = 1975.066 N/C

angle = 180-arctan(330.96/1947.14) = 180-9.64 =170.36 degrees from +x-axis counter clockwise direction

2)E = integral of k*lambda*dx/x^2 = integral of k*lambda0*x0*dx/x^3 = -k*lambda0*x0/2x^2

integrated from x0 to infinity it gives

E = k*lambda0*x0*(0-1/2x0^2 ) = k*lambda0/2x0

E = - k*lambda0/2x0 i N/C

bold letter represents vector notation and i is the unit vector in x-direction

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