Answer:
57.94°
Explanation:
we know that the expression of flux
where Ф= flux
E= electric field
S= surface area
θ = angle between the direction of electric field and normal to the surface.
we have Given Ф= 78 
E=
S=
= 
=0.5306
θ=57.94°
Answer:
it would be least to graetest
Explanation:
10-84
Answer:
E = k*Q₁/R₁² V/m
V = k*Q₁/R₁ Volt
Explanation:
Given:
- Charge distributed on the sphere is Q₁
- The radius of sphere is R₁
- The electric potential at infinity is 0
Find:
What is the electric field at the surface of the sphere?E.
What is the electric potential at the surface of the sphere?V
Solution:
- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.
- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F = k*Q₁/R₁²
- Then the electric field at that point is
E = F/1
E = k*Q₁/R₁² V/m
- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V = k*Q₁/R₁ Volt
