Respuesta :
Answer:
Part A : E = [tex]\frac{1}{4\pi}[/tex]ε₀ Q₁/R₁² Volt/meter
Part B : V = [tex]\frac{1}{4\pi}[/tex]ε₀ Q₁/R₁ Volt
Explanation:
Given that,
Charge distributed on the sphere is Q₁
The radius of sphere is R
₁
The electric potential at infinity is 0
Part A
The space around a charge in which its influence 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 = [tex]\frac{1}{4\pi}[/tex]ε₀ Q₁/R₁²
Then the electric field at that point is
E = F/1
E = [tex]\frac{1}{4\pi}[/tex]ε₀ Q₁/R₁² Volt/meter
Part B
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 = [tex]\frac{1}{4\pi}[/tex]ε₀ Q₁/R₁ Volt
(A) The electric field on the surface is [tex]E=\frac{1}{4\pi \epsilon_o}\frac{Q_1}{R_1^2}[/tex]
(B) The electric potential at the surface is [tex]V=\frac{1}{4\pi \epsilon_o}\frac{Q_1}{R_1}[/tex]
According to the question, there is a metal sphere with a radius [tex]R_1[/tex] and charge [tex]Q_1[/tex].
Electric field:
the charge on a metal sphere is spread on its surface since it is a conductor, there is no charge inside the sphere.
(a) The electric field E on the surface of the sphere is given by
[tex]E=\frac{1}{4\pi \epsilon_o}\frac{Q_1}{R_1^2}[/tex]
where [tex]\epsilon_o[/tex] is electrostatic constant
Electric potential:
(b) the potential difference is given by:
[tex]V=-\int\limits^{}_{} {E} \, dr\\\\V=-\frac{Q_1}{4\pi \epsilon_o}\int\limits^{\infty}_{R_1} {\frac{1}{r^2} } \, dr \\\\V=\frac{1}{4\pi \epsilon_o}\frac{Q_1}{R_1}[/tex]
Learn more about electric potential:
https://brainly.com/question/9383604?referrer=searchResults
