Please see below solution:
The electric field just outside its surface is
=
2
,
where is a radius of sphere, is a charge of sphere, =
1
4ε0
.
The electric field is
=
1
4ε0
2 =
1
ε0
∙
=
σ
ε0
,
where = 4
2
is an area of sphere, σ =
Q
S
is a surface charge density.
Answer:
.
20,000 because I did the quiz and got it right txt me if you need help
Answer:
0.53 N, 25.6°
Explanation:
side of triangle, a = 1.2 m
q = 7 μC
q1 = - 8 μC
q2 = - 6 μC
Let F1 be the force between q and q1
By using the coulomb's law
F1 = 0.35 N
Let F2 be the force between q and q2
By using the coulomb's law
F2 = 0.26 N
Write the forces in the vector form
Net force
Magnitude of the force
F = 0.53 N
Direction of force with x axis
θ = 25.6°
Resistance = (voltage) / (current) . . . <== MEMORIZE THIS
Resistance = (91 volts) / (10 Amperes)
Resistance = 9.1 ohms