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Question

Question: Find the energy contained in a cylinder of cross section \(10c{m^2}\) and length \(50cm\) along \(x ...

Find the energy contained in a cylinder of cross section 10cm210c{m^2} and length 50cm50cm along xx - axis, if E=50sinω(txc)E = 50\sin \omega \left( {t - \dfrac{x}{c}} \right) be the electric field in an electromagnetic wave.
(A) 5.5×1012J5.5 \times {10^{ - 12}}J
(B) 1.5×1011J1.5 \times {10^{ - 11}}J
(C) 6.2×1010J6.2 \times {10^{ - 10}}J
(D) 1.1×1015J1.1 \times {10^{ - 15}}J

Explanation

Solution

First convert all the quantities into SI units and then apply the energy formula to obtain an appropriate answer. In the formula, only the amplitude of the electric field should be used to calculate the energy.
U=12EDdV\Rightarrow U = \dfrac{1}{2}\int {\overrightarrow E } \cdot \overrightarrow D dV where E\overrightarrow E is the electric field vector and D\overrightarrow D is the displacement vector and VVis the finite region of space where the electric field is present.
D=ε0E\overrightarrow D = {\varepsilon _0}\overrightarrow E
U=12ε0E2dV\Rightarrow U = \dfrac{1}{2}\int {{\varepsilon _0}} {E^2}dV where VVis the finite region of space where the electric field is present.

Complete step by step answer:
Electrostatic energy is the energy associated with various electrostatic charge distributions. For such a system, the total energy is potential in nature and there is no kinetic energy.
We express the electrostatic energy in terms of field vectors E\overrightarrow E and D\Rightarrow \overrightarrow D where E\overrightarrow E is the electric field vector and D\Rightarrow \overrightarrow D is the displacement vector. The expression for the energy is given as,
U=12EDdV\Rightarrow U = \dfrac{1}{2}\int {\overrightarrow E } \cdot \overrightarrow D dV
Where VVis the finite region of space where the electric field is present.
Now, as we know the displacement vector D\overrightarrow D can be expressed in terms of electric field as given below,
D=ε0E\overrightarrow D = {\varepsilon _0}\overrightarrow E where VVis the finite region of space where the electric field is present.
Substituting this value in the energy expression we get,
U=12ε0E2dV\Rightarrow U = \dfrac{1}{2}\int {{\varepsilon _0}} {E^2}dV
We have to note that in this equation we only need to use the amplitude of the electric field E\overrightarrow E .
Before we start substituting the given values we need to convert all the quantities in SI unit as the answer is given in SI unit.
Therefore, area = 10×104m210 \times {10^{ - 4}}{m^2} and length = 50×102m50 \times {10^{ - 2}}m
The value of ε0=8.85×1012Fm1{\varepsilon _0} = 8.85 \times {10^{ - 12}}F{m^{ - 1}}
Substituting all the values given in the energy equation we get,
U=12×8.85×1012×502×10×104×50×102\Rightarrow U = \dfrac{1}{2} \times 8.85 \times {10^{ - 12}} \times {50^2} \times 10 \times {10^{ - 4}} \times 50 \times {10^{ - 2}}
U=5.5×1012J\Rightarrow U = 5.5 \times {10^{ - 12}}J

Therefore, the correct option is (A).

Note In this particular problem we assume the cylinder to be hollow and filled with air. However if we are given a problem where the cylinder is made up of any other material having a dielectric constant KK, then we replace ε0{\varepsilon _0} with ε=Kε0\varepsilon = K{\varepsilon _0} where ε\varepsilon is the permittivity of the given dielectric medium.