Respuesta :
Answer
given,
total charge Q = 100 n C
= 100 × 10⁻⁹ C
radius of the solid sphere = 30 cm
= 0.3 m
Volume of sphere = [tex]\dfrac{4}{3}\pi r^3[/tex]
= [tex]\dfrac{4}{3}\pi\times 0.3^3[/tex]
=0.113 m³
a) volume charge density
[tex]\rho = \dfrac{10^{-7}}{0.133}[/tex]
ρ = 8.85 × 10⁻⁷ C/m³
b) at r = 10 cm = 0.1 m
charge in the sphere at radius
[tex]Q = \dfrac{4}{3}\pi\times 0.1^3\time \rho[/tex]
= 3.7037 \times 10^{-9}C[/tex]
Field strength
[tex]E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}[/tex]
[tex]E_1 = \dfrac{3.7037 \times 10^{-9}}{4\pi \times 8.85\times 10^{-12}\times 0.1^2}[/tex]
= [tex]3.33 \times 10^3 N/C[/tex]
at r = 20 cm = 0.2 m
[tex]Q = \dfrac{4}{3}\pi\times r^3\time \rho[/tex]
[tex]E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}[/tex]
[tex]E_1 = \dfrac{ \dfrac{4}{3}\pi\times r^3\time \rho}{4\pi \epsilon_0 r^2}[/tex]
[tex]E_1 = \dfrac{\rho}{3 \epsilon_0}[/tex]
[tex]E_1 = \dfrac{0.2\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}[/tex]
= [tex]6.66 \times 10^3 N/C[/tex]
at r = 30 cm
[tex]E_1 = \dfrac{\rho}{3 \epsilon_0}[/tex]
[tex]E_1 = \dfrac{0.3\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}[/tex]
= 9.99 N/C
