
What will be the intensity of the electric field inside a uniformly charged conducting hollow sphere?
A). zero
B). non zero constant
C). change with $r$
D). inversely proportional to $r$.
Answer
457.8k+ views
Hint: Shell theorem is the basis for solving this question. The shell theorem is giving the simplifications in the gravitational matters that can be applicable in the case of bodies inside or outside a spherically symmetrical object. This theorem will have specific applications to astronomy. This will help you in answering this question.
Complete step-by-step solution
The shell theorem is to be applied here. According to shell’s theorem, the resultant force inside a shell will be,
$F\left( r \right)=\dfrac{GMm}{4{{r}^{2}}R}\int\limits_{R-r}^{R+r}{\left( \dfrac{1}{{{\left( {{s}^{p}} \right)}^{2}}}+\dfrac{{{r}^{2}}-{{R}^{2}}}{{{s}^{p}}} \right)}ds$
Where $M$ be the mass of the sphere, $m$ be the mass of the particle, $r$ is the distance between the particles and $R$ be the radius of the sphere.
This will be zero when $p=2$. This can help us to derive the electric field.
It says that a spherically symmetric object affects external bodies gravitationally even though all of its mass were concentrated at a specific position at its centre. When the body is a spherically symmetric shell, like a hollow sphere, there will be no resultant gravitational force acted by the shell on any of the bodies inside this without considering the location of the object within the shell. Therefore the intensity of the electric field inside a uniformly charged hollow sphere will be zero. And also we can say that outside the sphere, we can suppose that the sphere will be a point charge present at centre and can be calculated by the electric field. Therefore the voltage will also be a constant inside this hollow sphere as the electric field becomes zero.
Therefore the answer has been obtained. It has been mentioned as option A.
Note: The most famous scientist in physics, Sir Isaac Newton proposed this shell theorem. An electric field can be defined as the field that can be seen around each electric charge and will exert some force on all other charges in the field, this will be either attracting or repelling.
Complete step-by-step solution
The shell theorem is to be applied here. According to shell’s theorem, the resultant force inside a shell will be,

$F\left( r \right)=\dfrac{GMm}{4{{r}^{2}}R}\int\limits_{R-r}^{R+r}{\left( \dfrac{1}{{{\left( {{s}^{p}} \right)}^{2}}}+\dfrac{{{r}^{2}}-{{R}^{2}}}{{{s}^{p}}} \right)}ds$
Where $M$ be the mass of the sphere, $m$ be the mass of the particle, $r$ is the distance between the particles and $R$ be the radius of the sphere.
This will be zero when $p=2$. This can help us to derive the electric field.
It says that a spherically symmetric object affects external bodies gravitationally even though all of its mass were concentrated at a specific position at its centre. When the body is a spherically symmetric shell, like a hollow sphere, there will be no resultant gravitational force acted by the shell on any of the bodies inside this without considering the location of the object within the shell. Therefore the intensity of the electric field inside a uniformly charged hollow sphere will be zero. And also we can say that outside the sphere, we can suppose that the sphere will be a point charge present at centre and can be calculated by the electric field. Therefore the voltage will also be a constant inside this hollow sphere as the electric field becomes zero.
Therefore the answer has been obtained. It has been mentioned as option A.

Note: The most famous scientist in physics, Sir Isaac Newton proposed this shell theorem. An electric field can be defined as the field that can be seen around each electric charge and will exert some force on all other charges in the field, this will be either attracting or repelling.
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