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This work becomes the energy stored in the electrical field of the capacitor. In order to charge the capacitor to a charge Q , the total work required is

W = 0 W ( Q ) d W = 0 Q q C d q = 1 2 Q 2 C .

Since the geometry of the capacitor has not been specified, this equation holds for any type of capacitor. The total work W needed to charge a capacitor is the electrical potential energy U C stored in it, or U C = W . When the charge is expressed in coulombs, potential is expressed in volts, and the capacitance is expressed in farads, this relation gives the energy in joules.

Knowing that the energy stored in a capacitor is U C = Q 2 / ( 2 C ) , we can now find the energy density u E stored in a vacuum between the plates of a charged parallel-plate capacitor. We just have to divide U C by the volume Ad of space between its plates and take into account that for a parallel-plate capacitor, we have E = σ / ε 0 and C = ε 0 A / d . Therefore, we obtain

u E = U C A d = 1 2 Q 2 C 1 A d = 1 2 Q 2 ε 0 A / d 1 A d = 1 2 1 ε 0 ( Q A ) 2 = σ 2 2 ε 0 = ( E ε 0 ) 2 2 ε 0 = ε 0 2 E 2 .

We see that this expression for the density of energy stored in a parallel-plate capacitor is in accordance with the general relation expressed in [link] . We could repeat this calculation for either a spherical capacitor or a cylindrical capacitor—or other capacitors—and in all cases, we would end up with the general relation given by [link] .

Energy stored in a capacitor

Calculate the energy stored in the capacitor network in [link] (a) when the capacitors are fully charged and when the capacitances are C 1 = 12.0 μ F , C 2 = 2.0 μ F , and C 3 = 4.0 μ F, respectively.

Strategy

We use [link] to find the energy U 1 , U 2 , and U 3 stored in capacitors 1, 2, and 3, respectively. The total energy is the sum of all these energies.

Solution

We identify C 1 = 12.0 μ F and V 1 = 4.0 V , C 2 = 2.0 μ F and V 2 = 8.0 V , C 3 = 4.0 μ F and V 3 = 8.0 V . The energies stored in these capacitors are

U 1 = 1 2 C 1 V 1 2 = 1 2 ( 12.0 μ F ) ( 4.0 V ) 2 = 96 μ J, U 2 = 1 2 C 2 V 2 2 = 1 2 ( 2.0 μ F ) ( 8.0 V ) 2 = 64 μ J, U 3 = 1 2 C 3 V 3 2 = 1 2 ( 4.0 μ F ) ( 8.0 V ) 2 = 130 μ J .

The total energy stored in this network is

U C = U 1 + U 2 + U 3 = 96 μ J + 64 μ J + 130 μ J = 0.29 mJ .

Significance

We can verify this result by calculating the energy stored in the single 4 .0- μ F capacitor, which is found to be equivalent to the entire network. The voltage across the network is 12.0 V. The total energy obtained in this way agrees with our previously obtained result, U C = 1 2 C V 2 = 1 2 ( 4.0 μ F ) ( 12.0 V ) 2 = 0.29 mJ .

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Check Your Understanding The potential difference across a 5.0-pF capacitor is 0.40 V. (a) What is the energy stored in this capacitor? (b) The potential difference is now increased to 1.20 V. By what factor is the stored energy increased?

a. 4.0 × 10 −13 J ; b. 9 times

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In a cardiac emergency, a portable electronic device known as an automated external defibrillator (AED) can be a lifesaver. A defibrillator ( [link] ) delivers a large charge in a short burst, or a shock, to a person’s heart to correct abnormal heart rhythm (an arrhythmia). A heart attack can arise from the onset of fast, irregular beating of the heart—called cardiac or ventricular fibrillation. Applying a large shock of electrical energy can terminate the arrhythmia and allow the body’s natural pacemaker to resume its normal rhythm. Today, it is common for ambulances to carry AEDs. AEDs are also found in many public places. These are designed to be used by lay persons. The device automatically diagnoses the patient’s heart rhythm and then applies the shock with appropriate energy and waveform. CPR (cardiopulmonary resuscitation) is recommended in many cases before using a defibrillator.

Practice Key Terms 1

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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