Respuesta :
Answer:
(a). If z = 0, The electric field due to the rod is zero.
(b). If z = ∞, The electric field due to the rod is [tex]E\propto\dfrac{1}{z^2}[/tex].
(c). The positive distance is [tex]\dfrac{R}{\sqrt{2}}[/tex]
(d). The maximum magnitude of electric field is [tex]1.54\times10^{4}\ N/C[/tex]
Explanation:
Given that,
Radius = 2.00 cm
Charge = 4.00 mC
(a). If the radius and charge are R and Q.
We need to calculate the electric field due to the rod
Using formula of electric field
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}[/tex]
Where, Q = charge
z = distance
If z = 0,
Then, The electric field is
[tex]E=0[/tex]
(b). If z = ∞, z>>R
So, R = 0
Then, the electric field is
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}[/tex]
[tex]E\propto\dfrac{1}{z^2}[/tex]
(c). In terms of R,
We need to calculate the positive distance
If [tex]E\rightarrow E_{max}[/tex]
Then, [tex]\dfrac{dE}{dz}=0[/tex]
[tex]\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0[/tex]
Taking only positive distance
[tex]z=\dfrac{R}{\sqrt{2}}[/tex]
(d). If R = 2.00 and Q = 4.00 mC
We need to calculate the maximum magnitude of electric field
Using formula of electric field
[tex]E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}[/tex]
[tex]E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}[/tex]
[tex]E_{max}=15418.7\ N/C[/tex]
[tex]E_{max}=1.54\times10^{4}\ N/C[/tex]
Hence, (a). If z = 0, The electric field due to the rod is zero.
(b). If z = ∞, The electric field due to the rod is [tex]E\propto\dfrac{1}{z^2}[/tex].
(c). The positive distance is [tex]\dfrac{R}{\sqrt{2}}[/tex]
(d). The maximum magnitude of electric field is [tex]1.54\times10^{4}\ N/C[/tex]
