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BlackZzzverrR [31]
2 years ago
12

A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the

electric field due to this sphere at a point 5.0 cm outside the sphere's surface? (k=1/4πϵ0=8.99×109 N · m2/C2) A metal sphere of radius cm carries a charge of μC uniformly distributed over its surface. What is the magnitude of the electric field due to this sphere at a point cm outside the sphere's surface? ( N · m2/C2) 4.0×109 N/C 4.0×107 N/C 8.0×107 N/C 4.2×106 N/C 8.0×109 N/C
Physics
1 answer:
frozen [14]2 years ago
3 0

Answer:

8.0\cdot 10^5 N/C

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

E=k\frac{q}{(R+r)^2}

where

k=8.99\cdot 10^9 N m^2C^{-2} is the Coulomb's constant

q=2.0 \mu C=2.0 \cdot 10^{-6}C is the charge on the sphere

R=10 cm = 0.10 m is the radius of the sphere

r=5.0 cm = 0.05 m is the distance from the surface of the sphere

Substituting, we find

E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C

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0.68 m

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