
A parallel plate capacitor whose capacitance C is \[14pF\]is charged by a battery to a potential difference V=12V between its plates. The charging battery is now disconnected and a porcelain plate with k=7 is inserted between the plates, then the plate would oscillate back and forth between the plates, with a constant mechanical energy of ________pJ. (Assume no friction)
Answer
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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.
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