A hemispherical body is placed in a uniform electric field E. What is the flux linked with the curved surface, if the field is perpendicular to the base in figure.
$\begin{align}
& A.2\pi {{R}^{2}}E \\
& B.\dfrac{\pi {{R}^{2}}E}{2} \\
& C.\pi {{R}^{2}}E \\
& D.Zero \\
\end{align}$
Answer
648k+ views
Hint: The flux through a surface does not depend on its shape and size; it only depends upon the charge enclosed by the hemisphere. In this case the flux depends upon the electric field and the area of the cross section of the hemisphere.
Complete answer:
Values given to us are the uniform electric field E and the radius of the hemisphere R.
In this case, the flux entering the hemisphere is equal to the flux exiting it,
So,
$\begin{align}
& {{\phi }_{in}}={{\phi }_{out}} \\
& {{\phi }_{out}}=\pi {{R}^{2}}E \\
\end{align}$
So, the correct answer is “Option C”.
Additional Information:
Laws of physics like the Gauss's law for magnetism and Gauss's law for gravity have a great mathematical similarity with the Gauss law. Any inverse-square law can be deducted to Gauss's law. For example, Gauss's law itself is quite equivalent to the inverse-square Coulomb's law and Gauss's law for gravity is quite equivalent to the inverse-square Newton's law of gravity.
Note:
Electric field passing through a given area at a particular time is known as electric flux. Electric flux is directly proportional to the number of electric field lines passing through that given area. If the electric field is uniform, the electric flux passing through a surface of vector area S is: \[\phi =E\cdot S=EScos\theta \], where E is the magnitude of the electric field having units of V/m, S is the area of the surface, and θ is the angle between the electric field lines and the normal to S. Electric flux is also related to Gauss law. Gauss Law states the net electric flux from a closed surface is equal to $\dfrac{1}{{{\in }_{0}}}$ times the net charge enclosed in that area. There are two forms of the Gauss law; the integral form and the differential form.
Complete answer:
Values given to us are the uniform electric field E and the radius of the hemisphere R.
In this case, the flux entering the hemisphere is equal to the flux exiting it,
So,
$\begin{align}
& {{\phi }_{in}}={{\phi }_{out}} \\
& {{\phi }_{out}}=\pi {{R}^{2}}E \\
\end{align}$
So, the correct answer is “Option C”.
Additional Information:
Laws of physics like the Gauss's law for magnetism and Gauss's law for gravity have a great mathematical similarity with the Gauss law. Any inverse-square law can be deducted to Gauss's law. For example, Gauss's law itself is quite equivalent to the inverse-square Coulomb's law and Gauss's law for gravity is quite equivalent to the inverse-square Newton's law of gravity.
Note:
Electric field passing through a given area at a particular time is known as electric flux. Electric flux is directly proportional to the number of electric field lines passing through that given area. If the electric field is uniform, the electric flux passing through a surface of vector area S is: \[\phi =E\cdot S=EScos\theta \], where E is the magnitude of the electric field having units of V/m, S is the area of the surface, and θ is the angle between the electric field lines and the normal to S. Electric flux is also related to Gauss law. Gauss Law states the net electric flux from a closed surface is equal to $\dfrac{1}{{{\in }_{0}}}$ times the net charge enclosed in that area. There are two forms of the Gauss law; the integral form and the differential form.
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