Respuesta :
Explanation:
It is given that,
Charge, [tex]q=-6.5\ nC=-6.5\times 10^{-9}\ C[/tex]
Radius of the disk, R = 1.25 cm = 0.0125 m
(a) Electric field at a distance of 2 cm from the center of the disk. It can be calculated using the formula as :
[tex]E=\dfrac{\sigma}{2\epsilon_o}[1-\dfrac{x}{\sqrt{x^2+R^2}}][/tex]
Where
[tex]\sigma[/tex] is the surface charged density, [tex]\sigma=\dfrac{q}{\pi r^2}[/tex]
[tex]\sigma=\dfrac{-6.5\times 10^{-9}}{\pi (0.0125)^2}[/tex]
[tex]\sigma=-1.32\times 10^{-5}\ C/m^2[/tex]
So, [tex]E=\dfrac{1.32\times 10^{-5}}{2\times 8.85\times 10^{-12}}[1-\dfrac{0.02}{\sqrt{0.02^2+0.0125^2}}][/tex]
[tex]E=-1.13\times 10^5\ N/C[/tex]
The direction of electric field is outwards.
(b) Suppose that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. So, the electric field due to the ring is :
[tex]E=\dfrac{kqx}{(x^2+R^2)^{3/2}}[/tex]
[tex]E=\dfrac{9\times 10^9\times -6.5\times 10^{-9}\times 0.02}{(0.02^2+0.0125^2)^{3/2}}[/tex]
[tex]E=-8.91\times 10^4\ N/C[/tex]
The direction of electric field is outwards.
(c) If the charge is all brought to the center of the disk, The electric field at the center of disk is given by :
[tex]E=\dfrac{kq}{R^2}[/tex]
[tex]E=\dfrac{-9\times 10^9\times -6.5\times 10^{-9}}{0.02^2}[/tex]
[tex]E=-1.46\times 10^5\ N/C[/tex]
(d) he field in part (a) stronger than the field in part (b). This is due to the reason that the charge at the center of the ring is 0.
(e) The field in part (c) the strongest of the three fields because the for a point charge the charges are at same distances.
