In: Physics
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?
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 /r^{2}
dE = k dy /(x^{2} + y^{2})
In triangle AOB
Sin = y/r = y/sqrt(x^{2} + y^{2})
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 /(x^{2} + y^{2})) (y/sqrt(x^{2} + y^{2}))
dE' = 2 kydy / (x^{2} + y^{2})^{3/2}
Net electric field is given as
E = - dE'
E = (-2 k ) ydy / (x^{2} + y^{2})^{3/2}
E = (-2 k )
E = (- 2 x 9 x 10^{9} x 6 x 10^{-6})
x = 0.431
E = (-108000) ((1/sqrt((0.271)^{2} + 0.4^{2})) - (1/sqrt((0.271)^{2} + 0^{2}))
E = - 1.75 x 10^{5} N/C
So magnitude = 1.75 x 10^{5} N/C