Respuesta :
Answer:
Therefore,
a) [tex]E=20.82\ kN/C[/tex]
b) [tex]V= 2290\ Volt[/tex]
c) distance from the sphere's surface = 1.8 cm
Explanation:
Given:
Radius, r = 11 cm = 0.11 m
Charge, [tex]Q = 2.8\times 10^{-8}\ C[/tex]
To Find:
a) electric field at the sphere's surface = ?
b) If V = 0 at infinity, what is the electric potential at the sphere's surface = ?
c) At what distance from the sphere's surface has the electric potential decreased by 320 V = ?
Solution:
Electric field at the sphere's surface is given as,
[tex]E=\dfrac{k\times Q}{r^{2}}[/tex]
Where,
E = Electric Field,
[tex]k = Coulomb's\ constant = 9\times 10^{9}[/tex]
Q = Charge
r = Radius
Substituting the values we get
[tex]E=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{(0.11)^{2}}=20.82\ kN/C[/tex]
Now, Electric Potential at point surface is given as,
[tex]V=\dfrac{k\times Q}{r}[/tex]
Substituting the values we get
[tex]V=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{0.11}=2.29\ kV[/tex]
For distance from the sphere's surface has the electric potential decreased by 320 V,
So V becomes,
V = 2290 - 320 = 1970 Volt, then r =?
[tex]V=\dfrac{k\times Q}{r}[/tex]
Substituting the values we get
[tex]r=\dfrac{k\times Q}{V}\\\\r=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{1970}=0.128\ m[/tex]
Therefore the distance from the sphere's surface,
[tex]d = 12.8 - 11 = 1.8\ cm[/tex]
Therefore,
a) [tex]E=20.82\ kN/C[/tex]
b) [tex]V= 2290\ Volt[/tex]
c) distance from the sphere's surface = 1.8 cm
