CK-12-Physics - Intermediate

16.5. Electrical Energy Storage http://www.ck12.org

per unit charge, because like charges repel. This is analogous to stretching a spring. The farther the spring is

stretched, the more force is required to stretch it further. The total energy stored within the spring is equal to the

work done in stretching the spring. The workWis the product of the average forceFand displacement of the spring

x.

Analogously, the work done to charge the capacitor isW=F x= (QE)x=Q(Ex) =QV=Q

(

Vi+Vf

2

)

=^12 QVf. The

total charge placed on the capacitor isQand the initial voltageViof the capacitor is assumed to be zero.

The work can equally be expressed asW=^12 QV=^12 (CV)V=^12 CV^2 by substitutingQ=CVor asW=^12 QV^2 =^12 Q

2

C.

The subscript has been dropped on the voltage→V=Vf.

The potential energy stored within the capacitor is the same as the work done by the battery, that is,PEca p=^12 CV^2

or PEca p=^12 Q

2

C. The symbolPEis often replaced with the letterU→U=

1

2 CV

(^2) orU=^1
2
Q^2
C.
Check Your Understanding

1. The energy in a capacitor is 5.0 J and its capacitance is 4000μF. What is the potential difference across its plates?

Answer:U=^12 CV^2 =V^2 =^2 CU→V=

√

2 U

C =

√

2 ( 5. 0 J)

4000 × 10 −^6 f=^50 V

2. Two 1.0 mm gap capacitors,AandB, are separately charged by a 12-V battery. Each capacitor is removed from
   the battery after charging. The plates of capacitorBare pulled farther apart creating a 5.0 mm gap.
   Which of the following is correct?
   a. The charge and voltage of both capacitors remain the same.
   b. Only the charge on the capacitors is different.
   c. Only the voltage of the capacitors is different.
   d. The charge and voltage of both capacitors are different.
   Answer: The correct answer is C.
   The charges on the capacitors are the same since both capacitors were charged under identical conditions and then
   disconnected from the battery.
   Work, however, was done on capacitorB, since a force was applied (through a displacement) in order to increase the
   gap to 5 mm. The work increased the potential difference between the plates.
   Consider the equationQ=CV. If the distance between the plates is increased, then the capacitance decreases, since
   C=ε 0 Ad, in this cased→ 5 dsoC→^15 C. The voltageV′therefore must increaseQ=CV→^15 CV′→V′= 5 QC= 5 V.

Similarly, the electric fieldEremains constant, because we have two conducting parallel platesV=Ed→V′=

E( 5 d) = 5 Ed= 5 V.

1. The electric field between two parallel-plate conductors is considered uniform far away from the plate edges
   if the size of the plates is large compared to their separation distance.


2. The potential energy of a chargeqat a point between two parallel-plate conductors isPE=qEx, a reference
   point must be given such asPE=0 atx=0.


3. A point chargeqhas electric potential energyPExand electric potentialVxat pointx. Thus,PEx=qVx


4. The word voltage is used when we mean potential difference.


5. It is common to writeV=Ed, whereVis understood to mean the voltage (or potential difference) between
   the plates of a parallel-plate conductor anddis the distance between the plates.


6. The work done by the electric field in moving a charge between two parallel plate conductors isWf ield=
   −q∆V. The work done by an external force isWexternal f orce=q∆V.


7. Voltage can be thought of as the work per unit chargeV=Wq; that is, how much work is required per unit
   charge to move a charged particle in an electric field.