Respuesta :
Answer:
Potential at the surface of the conducting sphere: 1.1 × 10⁴ V.
Given that:
- Radius: 0.05 m;
- Charge density: 2.0 × 10⁻⁶ C/(m^2).
Explanation:
By applying Gauss Law, the electrical field outside a uniformly charged sphere is the same as if all the charge on it were concentrated at a point charge at the center of the sphere.
Surface area of this sphere:
[tex]4\pi\cdot r^{2} = 4\pi\times 0.05^{2} = \rm 0.0314159\; m^{2}[/tex].
Charge on the sphere:
[tex]\begin{aligned}&\rm \; 0.0314159\; m^{2}\times 2.0\times 10^{-6}\; C\cdot m^{-2}\\=&\; \rm 6.28319\times 10^{-8}\; C \end{aligned}[/tex].
Electrical potential at the surface of the sphere, as if the sphere is a point charge:
[tex]\begin{aligned}V &= \frac{k\cdot Q}{r} \\ &=\frac{8.99\times 10^{9}\times 6.28319\times 10^{-8}}{0.05}\\&\approx \rm 1.1\times 10^{4}\; V\end{aligned}[/tex].
The electric potential at the surface is [tex]1.1 \times 10^4 \rm \ V[/tex]. The electric flux through a closed surface is equal to the ratio of charge to the electric permittivity.
What does Gauss Law state?
The electric flux through a closed surface is equal to the ratio of charge to the electric permittivity.
[tex]V = \dfrac {k \times Q }{r }[/tex]
Where,
[tex]V[/tex] - electric potential
[tex]k [/tex]- electric constant = [tex]8.99 \times 10^{9} [/tex]
[tex]Q[/tex] - charge = [tex]6.283 \times 10^{-8} \rm \ C[/tex]
[tex]r[/tex] - distance = 0.05 m
Put the values in the formula,
[tex]V = \dfrac {8.99 \times 10^{9} \times 6.283 \times 10^{-8} \rm \ C }{ 0.05 \rm \ m}\\\\ V = 1.1 \times 10^4 \rm \ V[/tex]
Therefore, the electric potential at the surface is [tex]1.1 \times 10^4 \rm \ V[/tex].
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