Answer
Verified
99.9k+ views
Hint: As capacitor is used to store energy. To determine the amount of energy stored in a capacitor, we can use the parameters on which the energy stored depends upon and also their relations. Here also we use the formula for energy stored in a capacitor or capacitors.
Formula used:
The initial energy of capacitor,
\[{U_{}} = \dfrac{1}{2}C{V^2}\]
Where C is the capacitance and V is the potential difference
As, \[{\rm{ Q = CV}}\]
So, \[U = \dfrac{1}{2}\dfrac{{{Q^2}}}{{2kC}}\]
Where Q is charge on capacitor, K is dielectric constant and C is the capacitance.
Complete step by step solution:
Capacitance, C=\[14\,pF\]
Potential difference, V=12 V
Dielectric constant, k=7
The initial energy of capacitor,
\[{U_i} = \dfrac{1}{2}C{V^2}\]
\[\Rightarrow {U_i} = \dfrac{1}{2} \times 14 \times {(12)^2}\\
\Rightarrow {U_i} = 1008 pJ\]
The final energy of capacitor,
\[{U_f} = \dfrac{{{Q^2}}}{{2kC}}\]
\[\Rightarrow {U_f} = \dfrac{{{{\left( {14 \times 12} \right)}^2}}}{{2 \times 7 \times 14}}\\
\Rightarrow {U_f} = 144 pJ\]
Now the oscillating energy is,
\[{U_{osc}} = {U_i} - {U_f}\]
\[\Rightarrow {U_{osc}} = 1008 - 144\\
\therefore {U_{osc}} = 864 pJ\]
Therefore, the plate would oscillate back and forth between the plates, with a constant mechanical energy of 864 pJ.
Note: The capacitor is a device which stores electrical energy in the electrical field. A capacitor consists of two plates separated by a distance of equal and opposite charges. The area between the conductors may be filled by vacuum or insulating material known as dielectric. The work done to the charges from one plate to the other is stored as potential energy of the electric field of the conductor.
Formula used:
The initial energy of capacitor,
\[{U_{}} = \dfrac{1}{2}C{V^2}\]
Where C is the capacitance and V is the potential difference
As, \[{\rm{ Q = CV}}\]
So, \[U = \dfrac{1}{2}\dfrac{{{Q^2}}}{{2kC}}\]
Where Q is charge on capacitor, K is dielectric constant and C is the capacitance.
Complete step by step solution:
Capacitance, C=\[14\,pF\]
Potential difference, V=12 V
Dielectric constant, k=7
The initial energy of capacitor,
\[{U_i} = \dfrac{1}{2}C{V^2}\]
\[\Rightarrow {U_i} = \dfrac{1}{2} \times 14 \times {(12)^2}\\
\Rightarrow {U_i} = 1008 pJ\]
The final energy of capacitor,
\[{U_f} = \dfrac{{{Q^2}}}{{2kC}}\]
\[\Rightarrow {U_f} = \dfrac{{{{\left( {14 \times 12} \right)}^2}}}{{2 \times 7 \times 14}}\\
\Rightarrow {U_f} = 144 pJ\]
Now the oscillating energy is,
\[{U_{osc}} = {U_i} - {U_f}\]
\[\Rightarrow {U_{osc}} = 1008 - 144\\
\therefore {U_{osc}} = 864 pJ\]
Therefore, the plate would oscillate back and forth between the plates, with a constant mechanical energy of 864 pJ.
Note: The capacitor is a device which stores electrical energy in the electrical field. A capacitor consists of two plates separated by a distance of equal and opposite charges. The area between the conductors may be filled by vacuum or insulating material known as dielectric. The work done to the charges from one plate to the other is stored as potential energy of the electric field of the conductor.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
Other Pages
A series RLC circuit consists of an 8Omega resistor class 12 physics JEE_Main
The shape of XeF5 + ion is A Pentagonal B Octahedral class 11 chemistry JEE_Main
A block A slides over another block B which is placed class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Two billiard balls of the same size and mass are in class 11 physics JEE_Main
A tetracyanomethane B carbon dioxide C benzene and class 11 chemistry JEE_Main