Energy Store In A Capacitor
The energy stored in a parallel plate capacitor is electrostatic potential energy and is thus related to the charge Q and voltage V between the capacitor plates. A charged capacitor stores energy in the electric field between its plates. As the capacitor is being charged, the electric field builds up. When a charged capacitor is disconnected from a battery, its energy remains in the field in the space between its plates. The energy stored in a parallel plate capacitor is equal to half the product of the charge and the voltage, or half the product of the capacitance and the square of the voltage, as given by the formula:
\(\displaystyle U = \frac{1}{2}QV = \frac{1}{2}CV^2\)
Where U is the energy stored in the capacitor, Q is the charge on each plate, V is the potential difference between the plates, and C is the capacitance of the capacitor.
Let us consider a parallel plate capacitor with a charge +Q on one plate and -Q on the other, and a potential difference V between them. The capacitance of the capacitor is given by:
\(\displaystyle C = \frac{Q}{V}\)
The work done by the battery to charge the capacitor is equal to the energy supplied by the battery, which is the product of the charge and the voltage, i.e.,
\(\displaystyle W = QV\)
This energy is stored in the electric field between the plates of the capacitor. The electric field between the plates is uniform and perpendicular to the plates, and its magnitude is given by:
\(\displaystyle E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}\)
where \(\displaystyle \sigma\) is the surface charge density on one plate, \(\displaystyle\epsilon_0\) is the permittivity of free space, and A is the area of each plate. The energy density of the electric field is given by:
\(\displaystyle u_E = \frac{1}{2}\epsilon_0 E^2\)
If we multiply the energy density by the volume between the plates, we obtain the amount of energy stored between the plates of a parallel plate capacitor, i.e.,
\(\displaystyle U_C = u_E(Ad) = \frac{1}{2}\epsilon_0 E^2 Ad = \frac{1}{2}\epsilon_0 \frac{Q^2}{A^2\epsilon_0^2} Ad = \frac{1}{2}\frac{Q^2}{\epsilon_0 A}\)
where d is the separation between the plates. Using the formula for the capacitance, we can write the energy stored in the capacitor as:
\(\displaystyle U_C = \frac{1}{2}\frac{Q^2}{\epsilon_0 A} = \frac{1}{2}\frac{Q^2}{CV} = \frac{1}{2}\frac{QV}{V} = \frac{1}{2}QV\)
Alternatively, we can write the energy stored in the capacitor as:
\(\displaystyle U_C = \frac{1}{2}QV = \frac{1}{2}CV^2\)
These formulas show that the energy stored in a parallel plate capacitor is equal to half the product of the charge and the voltage, or half the product of the capacitance and the square of the voltage.
Energy Density In Electric Field
The energy density in an electric field is the amount of energy stored per unit volume in the region of space where the electric field exists. It is a measure of how much work can be done by the electric field on a unit charge. The energy density in an electric field is given by the formula:
\(\displaystyle u_E = \frac{1}{2}\epsilon_0 E^2\)
where \(\displaystyle u_E\) is the energy density, \(\displaystyle\epsilon_0\) is the permittivity of free space, and E is the magnitude of the electric field. This formula can be derived from the formula for the energy stored in a parallel plate capacitor, as follows:
The energy stored in a parallel plate capacitor is equal to half the product of the charge and the voltage, or half the product of the capacitance and the square of the voltage, as given by the formula:
\(\displaystyle U_C = \frac{1}{2}QV = \frac{1}{2}CV^2\)
where \(\displaystyle U_C\) is the energy stored in the capacitor, Q is the charge on each plate, V is the potential difference between the plates, and C is the capacitance of the capacitor. The capacitance of a parallel plate capacitor is given by:
\(\displaystyle C = k \epsilon_0 \frac{A}{d}\)
where k is the relative permittivity of the dielectric material, \(\displaystyle\epsilon_0\) is the permittivity of free space, A is the area of each plate, and d is the distance between them. The electric field between the plates is uniform and perpendicular to the plates, and its magnitude is given by:
\(\displaystyle E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}\)
where \(\displaystyle\sigma\) is the surface charge density on one plate. Substituting these expressions into the formula for the energy stored in the capacitor, we get:
\(\displaystyle U_C = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{1}{2}k \epsilon_0 \frac{A}{d}V^2 = \frac{1}{2}k \epsilon_0 \frac{A}{d}\left(\frac{Ed}{k}\right)^2 = \frac{1}{2}\epsilon_0 E^2 Ad\)
∵ \(\displaystyle \left( {C=k{{\varepsilon }_{0}}\frac{A}{d}} \right)\)
where we have used the fact that \(\displaystyle V = Ed/k\) comes from the definition of electric potential and the electric field of a parallel plate capacitor. Dividing both sides by the volume Ad, we obtain the energy density in the electric field, i.e.,
\(\displaystyle u_E = \frac{U_C}{Ad} = \frac{1}{2}\epsilon_0 E^2\)
This formula shows that the energy density in an electric field is proportional to the square of the electric field and the permittivity of free space.
Common Potential During System Of Charges
Let us consider two capacitors \(\displaystyle C_1\) and \(\displaystyle C_2\) with different charges \(\displaystyle Q_1\) and \(\displaystyle Q_2\) and different voltages \(\displaystyle V_1\) and \(\displaystyle V_2\). When they are connected in parallel, they start sharing their charges until they reach a common potential V. The total charge of the system is conserved, i.e.,
\(\displaystyle Q = Q_1 + Q_2\)
where Q is the total charge. The equivalent capacitance of the parallel combination is the sum of the individual capacitances, i.e.,
\(\displaystyle C_{eq} = C_1 + C_2\)
where \(\displaystyle C_{eq}\) is the equivalent capacitance. The common potential is defined as the ratio of the total charge to the equivalent capacitance, i.e.,
\(\displaystyle V = \frac{Q}{C_{eq}}\)
Substituting the expressions for Q and \(\displaystyle C_{eq}\), we get:
\(\displaystyle V = \frac{Q_1 + Q_2}{C_1 + C_2}\)
Using the relation \(\displaystyle Q_i = C_i V_i\) for each capacitor, we can write the common potential as:
\(\displaystyle V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}\)
This is the formula for the common potential of capacitors during the sharing of charges. It shows that the common potential is a weighted average of the initial voltages of the capacitors, with the weights being their capacitances.
Loss Of Energy During Sharing Of Charges
The initial energy of the system is the sum of the energies stored in the two capacitors before they are connected, i.e.,
\(\displaystyle U = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2\)
where U is the initial energy, \(\displaystyle C_1\) and \(\displaystyle C_2\) are the capacitances of the two capacitors, and \(\displaystyle V_1\) and \(V_2\) are their initial voltages. The final energy of the system is the energy stored in the equivalent capacitor after they are connected, i.e.,
\(\displaystyle U’ = \frac{1}{2} C_{eq} V^2\)
where U’ is the final energy, \(\displaystyle C_{eq}\) is the equivalent capacitance of the parallel combination, and V is the common potential. The equivalent capacitance and the common potential are given by the formulas:
\(\displaystyle C_{eq} = C_1 + C_2\)
\(\displaystyle V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}\)
The energy loss is the difference between the initial and final energies, i.e.,
\(\displaystyle\Delta U = U – U’ = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 – \frac{1}{2} C_{eq} V^2\)
Substituting the expressions for \(\displaystyle C_{eq}\) and $V$, we get:
\(\displaystyle\Delta U = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 – \frac{1}{2} (C_1 + C_2) \left(\frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}\right)^2\)
Expanding and simplifying, we get:
\(\displaystyle\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 – V_2)^2\)
This equation shows that the energy loss is proportional to the square of the difference between the initial voltages of the two capacitors.