Answer:
Electric field by charged disk is given as
E = (Charge Density/2u0)*[1 - (z/sqrt(z^2 - R^2))]
R = 9.54cm = 0.0954m, z = 1.01m, Charge density = 4.07 x 10^-6C/m2, e0 = 8.85 x 10^-12F/m.
Substituting all the values in to equation,
E = (2.299 x 10^5) x (8.931 x 10^-3)
E = 2.053 x 10^3N/C
Explanation:
The electric field produced by this disk at point P is [tex]\bold {2.053 x 10^3\ N/C}[/tex] along the x-axis.
Electric field by the charged disk is given as
[tex]E = k \sigma 2 \pi\times[ 1 - \dfrac z { \sqrt {z^2 - R^2}}][/tex]
Where,
[tex]\sigma[/tex]- charge dencity = [tex]4.07 \times 10^{-6}\rm \ C/m^2[/tex]
[tex]z[/tex] - distance of point = 1.01 m
[tex]R[/tex] - radius = 9.54cm = 0.0954 m
[tex]k[/tex] - Coulomb's constant = [tex]8.99 \times 10^9 \rm \ Nm[/tex]
Put the values in the equation,
[tex]E = 8.99 \times 10^9 \times 2 \pi\times[ 1 - \dfrac {1.01} { \sqrt {(1.01)^2 - (0.0954)^2}}][/tex]
[tex]\bold {E = 2.053 x 10^3\ N/C}[/tex]
Therefore, the electric field produced by this disk at point P is[tex]\bold {2.053 x 10^3\ N/C}[/tex] along the x-axis.
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