Question

Continuous charge distribution

Answer 1

E(x)=-integral dEcos(theta)=-integral kedq*cos(theta)=-ke(lambda) integral xdx/(x^2+d^2)^3/2
let x=dtan(theta) dx=dsec^2(theta)d(theta)
E(x)=-ke(lambda)/dintegral sin(theta)
I set my limits to theta maximum and zero
E(x)=ke(lambda)/d (cos(theta maximum)-1)

Answer 2

Cos(theta maximum)=L/(L^2+d^2)^1/2

Answer 3

E(x)=keQ/L*d((L/L^2+d^2)^1/2-1)

Answer 4

the infinitesimal electric field, is given by the subsequent vector:
\[\huge d{\mathbf{E}} = \frac{{\lambda dx}}{{{{\left( {{x^2} + {d^2}} \right)}^{3/2}}}}\left( { - x,d} \right)\]
where:
\[\huge \lambda = \frac{Q}{L}\]
is the charge linear density
\[\large {E_x} = \frac{Q}{{Ld}}\left( {\frac{d}{{\sqrt {{L^2} + {d^2}} }} - 1} \right) = \frac{Q}{{Ld}}\left( {\frac{1}{{\sqrt {1 + {{\left( {\frac{L}{d}} \right)}^2}} }} - 1} \right)\]
next I use this approximation:
\[\Large \frac{1}{{\sqrt {1 + {{\left( {\frac{L}{d}} \right)}^2}} }} \simeq 1 - \frac{{{L^2}}}{{2{d^2}}}\]
which comes from the Taylor expansion
So, I get:
\[\huge {E_x} \simeq - \frac{{QL}}{{2{d^3}}}\]

Answer 5

next, we have:
\[\huge {E_y} = \frac{Q}{{Ld}}\frac{{L/d}}{{\sqrt {1 + {{\left( {\frac{L}{d}} \right)}^2}} }}\]
again I use this approximation:
\[\Large \frac{1}{{\sqrt {1 + {{\left( {\frac{L}{d}} \right)}^2}} }} \simeq 1 - \frac{{{L^2}}}{{2{d^2}}}\]
so, I get:
\[\Large \begin{gathered}
{E_y} = \frac{Q}{{Ld}}\frac{{L/d}}{{\sqrt {1 + {{\left( {\frac{L}{d}} \right)}^2}} }} = \frac{Q}{{{d^2}}}\frac{1}{{\sqrt {1 + {{\left( {\frac{L}{d}} \right)}^2}} }} \simeq \hfill \\
\hfill \\
\simeq \frac{Q}{{{d^2}}}\left( {1 - \frac{{{L^2}}}{{2{d^2}}}} \right) = \frac{Q}{{{d^2}}} - \frac{{Q{L^2}}}{{2{d^4}}} \simeq \frac{Q}{{{d^2}}} \hfill \\
\end{gathered} \]

Answer 6

The electric field in the x direction is given as Q/Ld(d/(d^2+L^2)^0.5)-1)
When I proceeded on with the integration and set the correct limits my result was different instead of the d was L. Shouldnt it be the same?