Answer:
The electric field inside the hollow plastic ball is zero.
Explanation:
According to Gauss's law, the electric field at the closed Gaussian surface [tex]S[/tex] is given by
[tex]$\oint_S {E_n \cdot dA = \dfrac{Q_{enc}}{{\varepsilon _0 }}}$[/tex]
Now, to find the electric field inside the hollow plastic ball, we choose a spherical Gaussian surface inside the ball. And since all of the charge lies on the surface of the ball, the Gaussian surface does not enclose any charge; therefore, the Gauss's law gives:
[tex]$ E(4\pi r^2)= \dfrac{Q_{enc}}{{\varepsilon _0 }}}$[/tex]
[tex]$ E(4\pi r^2)=0$\\\\\boxed{E = 0}[/tex]
The electric field inside the hollow plastic ball is zero.
The electric field inside the hollow plastic ball is zero.
It is given that the charge is distributed on the surface of the ball, there is no charge inside the ball.
According to the Gauss Law, the electric flux out of a closed surface is
Q/ε₀ times the total charge enclosed by the surface. Mathematically the electric flux φ is the product of Electric field E and surface area A, it is expressed as :
[tex]\phi=E.A\\\\\frac{Q_{enclosed}}{\epsilon_0}=E.A[/tex]
Inside the hollow plastic ball, the charge enclosed [tex]Q_{enclosed}=0[/tex]
Therefore, electric field E = 0.
Learn more about Gauss Law:
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