Question

a solid metal sphere with radius 0.450 m carries a net charge of 0.250 nc. find the magnitude of the electric field (a) at a point 0.100 m outside the surface of the sphere and (b) at a point inside the sphere, 0.100 m below the surface.

Answer 1

To find the electric field at a point outside the surface of the sphere, use the formula E = k * (Q / r²). For part (a), the magnitude of the electric field at a point outside the surface of the sphere is approximately 1.168 x 10⁶ N/C. For part (b), the electric field inside a uniformly charged sphere is zero, so the magnitude of the electric field at a point inside the sphere, 0.100 m below the surface, is zero.

Step-by-step explanation:

To find the electric field at a point outside the surface of the sphere, we can use the formula:

E = k * (Q / r²)

where E is the electric field, k is the electrostatic constant, Q is the net charge on the sphere, and r is the distance from the center of the sphere to the point.

For part (a), we have Q = 0.250 nC and r = 0.100 m + 0.450 m = 0.550 m. Plugging these values into the formula, we get:

E = (9.0 x 10⁹ Nm²/C²) * (0.250 x 10⁻⁹ C) / (0.550 m)²

Solving this equation, we find that the magnitude of the electric field at the point outside the surface of the sphere is approximately 1.168 x 10⁶ N/C.

For part (b), the electric field inside a uniformly charged sphere is zero. Therefore, the magnitude of the electric field at a point inside the sphere, 0.100 m below the surface, is zero.

Answer 2

a) E = K Q / R^2 because of symmetry all of the charge on the sphere will act as a charge at the center of the sphere

E = 9.00E9 * .25E-9 / .55^2 = 7.44 N/C

b) there can be no electric field within the sphere else a current would flow

E = 0