Respuesta :
Answer:
a, V = 65.6V
b. V = 131.0V
C. V = 131.0V
Explanation:
Given
Total electric charge [tex]q= 3.50nC=3.50\times10^{-9}C[/tex] distributed uniform at surface of metal sphere where radius is R = 24cm
R = 24cm = 0.024m
we have to find potential at different distance
a. r = 48cm = 0.048m
b. r = 24cm = 0.024m
c. r = 12cm = 0.012m
a. solution
Here the distance r is grater then R,we can get the potential outside of sphere
[tex]V = \dfrac{q}{4\pi\varepsilon_0r}[/tex]
[tex]\textrm{where is} \dfrac{1}{4\pi\varepsilon_0}=9.0\times10^{9}N.m^{2}/c^2[/tex]
now we plug the value q and r in potential v
[tex]V=(9\times10^9N.m^2/C^2)\dfrac{3.50\times10^{-9}C}{0.48m}[/tex]
=65.6V
b.solution
Here distane r is the same from radius R, so we can get potential inside sphere
[tex]v= \dfrac{q}{4\pi\varepsilon_0R}[/tex]
Now we plug the value of R and q
[tex]V=(9\times10^9N.m^2/C^2)\dfrac{3.50\times10^{-9}C}{0.24m}[/tex]
[tex]V=131V[/tex]
3.solution
Inside sphere electric feild is zero.so potential is the same every point inside sphere and equal potential on the surface
[tex]V=131V[/tex]
(a) The potential at the distance of 48.0 cm from the centre of the uniformly charged sphere is 65.625 J/C
(b) The potential at the distance of 24.0 cm from the centre of the sphere is 112.5 J/C
(c) The potential at the distance of 12.0 cm from the centre of the sphere is 112.5 J/C
The answer is explained as given below.
Electrostatic Potential
(a) The potential at a point 'r' from the centre of a uniformly charged sphere, where r > R, is given by;
Here, [tex]r=48\,cm=0.48\,m[/tex]
- [tex]V=k\frac{Q}{r}[/tex]
- Where [tex]k=9\times 10^9 Nm^2/C^2[/tex] is the Coulomb constant and Q is the total electric charge of the sphere.
- [tex]Q=3.50\times 10^{-9}\,C[/tex]
Substituting the given values, we get;
[tex]V=(9\times 10^9\,Nm^2/C^2)\times\frac{3.50\times 10^{-9}\,C}{0.48\,m} =65.625\,J/C[/tex]
(b) The electric field inside a conducting sphere is zero.
So, the potential inside a charged sphere is constant and is given by;
[tex]V_{inside}= K\frac{q}{R}[/tex]
Where R is the radius of the sphere.
So, [tex]V= (9\times 10^9 Nm^2/C^2) \times \frac{3.50\times 10^{-9}C}{0.24\,m}=112.5\,J/C[/tex]
(c) Since the given distance here is less than the radius of the sphere, here also the potential is given by,
[tex]V= (9\times 10^9 Nm^2/C^2) \times \frac{3.50\times 10^{-9}C}{0.24\,m}=112.5\,J/C[/tex]
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