Answer:
[tex]2.62898\times 10^{-6}\ C/m^3[/tex]
[tex]1979.99974\ N/C[/tex]
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
Q = Charge
r = Distance = 8 cm
R = Radius = 4 cm
Electric field is given by
[tex]E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C[/tex]
Volume charge density is given by
[tex]\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3[/tex]
The volume charge density for the sphere is [tex]2.62898\times 10^{-6}\ C/m^3[/tex]
[tex]E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C[/tex]
The magnitude of the electric field is [tex]1979.99974\ N/C[/tex]
The volume charge density for the sphere and the magnitude of the electric field is mathematically given as
a)[tex]\phi=2.683-6[/tex]
b)E=1979.9997N/C
Generally the equation for the electric field is mathematically given as
[tex]E=\frac{KQ}{r^2}[/tex]
Therefore
[tex]Q=\frac{Er^2}{k}[/tex]
Q=7.047e-10
Therefore
[tex]\phi=\frac{7.047e-10}{4/3\pi(0.04)^3}\\\\[/tex]
[tex]\phi=2.683-6[/tex]
Hence,The magnitude of the electric field is
[tex]E=\frac{kQr}{R^3}\\\\E=\frac{8.99e9*7.047e-10*0.02}{0.04^3}[/tex]
E=1979.9997N/C
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