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Gre4nikov [31]
3 years ago
14

uniformly charged conducting sphere of 1.2 m diameter has surface charge density 8.1 mC/m2. Find (a) the net charge on the spher

e and (b) the total electric flux leaving the surface.
Physics
1 answer:
Stels [109]3 years ago
5 0

Answer:

(a) = 3.7 × 10⁻⁵

(b) = 4.1 × 10⁻⁶ N.M²/C

Explanation:

(a) Diameter of the sphere, d = 1.2 m

Radius of the sphere, r = 0.6 m

Surface charge density, = 8.1 mC/m2 = 8.1 × 10⁻⁶ C/m²

Total charge on the surface of the sphere,

Q = Charge density × Surface area

    = 4πr²σ

   = 4 (3.14) (0.12²) (8.1 × 10⁻⁶)

   = 3.66 × 10⁻⁵C

   ≅ 3.7 × 10⁻⁵C

Therefore, the net charge on the sphere is  3.7 × 10⁻⁵C

(b)

Total electric flux (∅)

=Q / ε₀

ε₀ =  8.854 × 10⁻¹² N⁻¹C² m⁻²

Q =   3.66 × 10⁻⁵C

 =  3.66 × 10⁻⁵ / 8.854 × 10⁻¹²

= 4.1 × 10⁻⁶ N.M²/C

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Answer:

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\theta = Angle = 26^{\circ}

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2

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v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m

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Explanation:

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See attached file

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