Answer:
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:
where
is the Coulomb's constant
is the charge on the sphere
is the radius of the sphere
is the distance from the surface of the sphere
Substituting, we find
Answer:
(a) 0.0178 Ω
(b) 3.4 A
(c) 6.4 x 10⁵ A/m²
(d) 9.01 x 10⁻³ V/m
Explanation:
(a)
σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹
d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) π d²
A = (0.25) (3.14) (2.6 x 10⁻³)²
A = 5.3 x 10⁻⁶ m²
L = length of the wire = 6.7 m
Resistance of the wire is given as
R = 0.0178 Ω
(b)
V = potential drop across the ends of wire = 0.060 volts
i = current flowing in the wire
Using ohm's law, current flowing is given as
i = 3.4 A
(c)
Current density is given as
J = 6.4 x 10⁵ A/m²
(d)
Magnitude of electric field is given as
E = 9.01 x 10⁻³ V/m