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Question: A hollow metal sphere of radius \(R\) is uniformly charged. The electric field due to the sphere at ...

A hollow metal sphere of radius RR is uniformly charged. The electric field due to the sphere at a distance rr from the center.
A) Increases as rr increases for r<Rr < R and for r>Rr > R .
B) Zero as rr increases for r<Rr < R , decreases as rr increases for r>Rr > R .
C) Zero as rr increases for r<Rr < R , increases as rr increases for r>Rr > R .
D) Decreases as rr increases for r<Rr < R and for r>Rr > R .

Explanation

Solution

We can use Gauss Law to solve this problem, because Gauss Law can be stated using the electric field. Charge is distributed over the surface of a hollow metal sphere. Take both the hollow metal sphere into consideration, i.e., inside, and outside, for getting the correct answer.
Formula used:
ϕe=E.dA=Qεo{{\phi }_{e}}=\oint{\overrightarrow{E}.\overrightarrow{dA}=\dfrac{Q}{{{\varepsilon }_{o}}}}, where EE is the electric field,
dAdA is a vector representing an infinitesimal element of area of the surface,
QQ is the total charge enclosed within volume, and
εo{{\varepsilon }_{o}} is the electric constant.

Complete answer:


According to Gauss Law, the net flux through any closed surface equals the total charge inside that surface divided by εo{{\varepsilon }_{o}} . The equation for this law can be written as:
ϕe=E.dA=Qnetεo{{\phi }_{e}}=\oint{\overrightarrow{E}.\overrightarrow{dA}=\dfrac{{{Q}_{net}}}{{{\varepsilon }_{o}}}}
For rBy Gauss law, $${{\phi }_{e}}=\oint{{{\overrightarrow{E}}_{in}}.\overrightarrow{dA}=\dfrac{{{Q}_{net}}}{{{\varepsilon }_{o}}}=0\,(\because {{Q}_{net}}=0)}$$ \Rightarrow {{E}{in}}=0For Forr>R(outside sphere): By Gauss Law, $${{\phi }_{e}}=\oint{{{\overrightarrow{E}}_{o}}.\overrightarrow{dA}=\dfrac{{{q}_{en}}}{{{\varepsilon }_{o}}}}$$ Here,{{q}{en}}=Q,and,dA=4\pi {{r}^{2}}Substitutethevalueintheaboveequation,weget: Substitute the value in the above equation, we get: \begin{aligned}
& \Rightarrow {{E}{o}}\times 4\pi {{r}^{2}}=\dfrac{Q}{{{\varepsilon }{o}}} \\
& \Rightarrow {{E}_{o}}\propto \dfrac{1}{{{r}^{2}}} \\
\end{aligned}Fromtheabovesolution,itisclearthattheelectricfieldiszeroas From the above solution, it is clear that the electric field is zero asrincreasesforincreases forr < Randdecreasesasand decreases asrincreasesforincreases forr > R$.

Therefore, the correct answer is Option (B).

Note:
Substitute the value properly in both the condition to get the final relation. It is necessary to derive both the relations before marking the result, so that we can get the dependency of the electric field, which will help to reach a solution.