Answer:
a)[tex]\rho = 8.9\times 10^{-7}\ C/m^2[/tex]
b)E=101,166.9 N/C
Explanation:
Given that
Diameter ,d=0.98 m
Charge ,q=2.7μC
a)
Surface charge density(ρ)
Surface charge density is the charge per unit surface area of sphere.
So
[tex]\rho =\dfrac{q}{\pi d^2}\ C/m^2[/tex]
[tex]\rho =\dfrac{2.7\times 10^{-6}}{\pi \times 0.98^2}\ C/m^2[/tex]
[tex]\rho = 8.9\times 10^{-7}\ C/m^2[/tex]
b)
Electric field E
[tex]E =\dfrac{\rho}{\varepsilon _o}\ N/C[/tex]
[tex]E =\dfrac{8.9\times 10^{-7}}{8.85\times 10^{-12}}\ N/C[/tex]
E=101,166.9 N/C
