Question

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

Answer 1

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

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

Answer 2

The magnitude of the electric field due to this sphere at the given distance is 8 x 10⁵ N/C.

The given parameters;

• Radius of the sphere, r = 10 cm = 0.1 m
• Charge of the sphere, Q = 2 μC = 2 x 10⁻⁶ C
• Distance outside the sphere, x = 5 cm = 0.05 m

The magnitude of the electric field due to this sphere at the given distance is calculated using Coulomb's law;

$E = \frac{kQ}{R^2}$

where;

• R is the total distance from the center to the external distance of the sphere;

R = (0.1 + 0.05) m = 0.15 m

$E = \frac{8.99\times 10^{9} \times 2\times 10^{-6}}{(0.15)^2} \\\\ E= 8 \times 10^{5} \ N/C$

Thus, the magnitude of the electric field due to this sphere at the given distance is 8 x 10⁵ N/C.

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