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melomori [17]
4 years ago
10

A solid sphere of radius 30cm is uniformly charged to 100nC. a) What is the volume charge density of the sphere? b) What is the

strength of the E-field at r-10cm, r-20cm, and r-30cm
Physics
1 answer:
g100num [7]4 years ago
6 0

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 = \dfrac{4}{3}\pi r^3

                              = \dfrac{4}{3}\pi\times 0.3^3

                              =0.113 m³

a) volume charge density

\rho = \dfrac{10^{-7}}{0.133}

         ρ  = 8.85 × 10⁻⁷ C/m³

b) at r = 10 cm = 0.1 m

charge in the sphere at radius

Q = \dfrac{4}{3}\pi\times 0.1^3\time \rho

   = 3.7037 \times 10^{-9}C[/tex]

Field strength

E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}

E_1 = \dfrac{3.7037 \times 10^{-9}}{4\pi \times 8.85\times 10^{-12}\times 0.1^2}

      = 3.33 \times 10^3 N/C

at r = 20 cm = 0.2 m

Q = \dfrac{4}{3}\pi\times r^3\time \rho

E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}

E_1 = \dfrac{ \dfrac{4}{3}\pi\times r^3\time \rho}{4\pi \epsilon_0 r^2}

E_1 = \dfrac{\rho}{3 \epsilon_0}

E_1 = \dfrac{0.2\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}

      = 6.66 \times 10^3 N/C

at r = 30 cm

E_1 = \dfrac{\rho}{3 \epsilon_0}

E_1 = \dfrac{0.3\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}

      = 9.99 N/C

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