Respuesta :
Answer:
(a) [tex]E = -1.02 \times 10^5~N/C[/tex]
(b) [tex]E = -9.7 \times 10^4~N/C[/tex]
Explanation:
(a) The electric field for a point charge is given by the following formula:
[tex]\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r[/tex]
Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.
[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}[/tex]
Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.
In cylindrical coordinates, da = r dr dθ. So,
[tex]\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}[/tex]
Hence, dE is now:
[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}[/tex]
The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.
It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.
Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,
[tex]\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}[/tex]
Therefore, vertical component of dE now becomes
[tex]dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})[/tex]
Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:
[tex]E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C[/tex]
(b) We will have a similar approach, but a simpler integral.
[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi[/tex]
[tex]E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C[/tex]
Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.
(a) The electric field is [tex]E =-1.02*10^5N/C[/tex] toward the center of the disc.
(b) The electric field is [tex]E=-9.7*10^4N/C[/tex]
To calculate the electric field intensities and direction we need to apply the equation for the electric field above the axis passing through the center of a disc and a ring as given below:
(a) The electric field due to the uniformly charged disk
The electric field E due to uniformly charged disk along its axis is:
[tex]E_z=\frac{Qz}{2\pi \epsilon_oR^2} (\frac{1}{z}-\frac{1}{\sqrt{z^2+R^2} })[/tex]
where, R is radius of disk = 1.05 cm = 1.05×10⁻²m
z is the distance along the axis = 2.5cm = 2.5×10⁻²m
Q = -8nC = -8×10⁻⁹C
[tex]E_z=\frac{-8*10^{-9}*2.5*10^{-2}*2*9*10^9}{(1.05*10^{-2})^2} (\frac{1}{2.5*10^{-2}}-\frac{1}{\sqrt{(2.5*10^{-2})^2+(1.05*10^{-2})^2} })\\\\E_z=-1.02*10^5N/C[/tex]
the negative sign indicates that the field is directed towards the center of the disc.
(b)The electric field due to the uniformly charged ring
The electric field due to uniformly charged ring along its axis is:
[tex]E_z=\frac{Qz}{4\pi \epsilon_o(z^2+R^2)^{3/2}}[/tex]
where, R is radius of disk = 1.05 cm = 1.05×10⁻²m
z is the distance along the axis = 2.5cm = 2.5×10⁻²m
Q = -8nC = -8×10⁻⁹C
[tex]E_z=\frac{-8*10^{-9}*2.5*10^{-2}*9*10^9}{[(2.5*10^{-2})^2+(1.05*10^{-2})^2]^{3/2}} \\\\E_z=-9.7*10^4N/C[/tex]
the negative sign indicates that the field is directed towards the center of the ring.
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