Electricity and Magnetism apacitance
Sources of Electric Potential A potential difference can be created by moving charge from one conductor to another. The potential difference on a capacitor can produce a current (flow of charge), but this current cannot be sustained because the charge separation and potential difference rapidly disappears. 2
Forming a apacitor Any two conducting electrodes can form a capacitor, regardless of their shape. Q V The capacitance depends only on the geometry of the electrodes, not on their present charge or potential difference. (In fact, one of the electrodes can be moved to infinity, so the capacitance of a single electrode is a meaningful concept.) 3
In this chapter we will cover the following topics: -apacitance of a system of two isolated conductors. -alculation of the capacitance for parallel plate capacitor (will return and do more later). -Methods of connecting capacitors (in series, in parallel). -Equivalent capacitance. -Energy stored in a capacitor (will do later). -Behavior of an insulator (a.k.a. dielectric) when placed in the electric field created in the space between the plates of a capacitor (will look at later). 4
Parallel Plate apacitor A parallel plate capacitor is defined as made up from two parallel plane plates of area A separated by a distance d. The electric field between the plates and away from the plate edges is uniform. lose to the plates edges the electric field (known as "fringing field") becomes non-uniform. V + - _ Batteries A battery is a device that maintains a constant potential difference V between its two terminals. These are indicated in the battery symbol using two parallel lines unequal in length. The longer line indicates the terminal at higher potential while the shorter line denotes the lower potential terminal. 5
harging a apacitor One method to charge a capacitor is shown in the figure. When the switch S is closed, the electric field of the battery drives electrons from the battery negative terminal to the capacitor plate connected to it (labeled " l" for low). The battery positive terminal removes an equal number of electrons from the plate connected to it (labeled " h" for high). Initially the potential difference V between the capacitor plates is zero. The charge on the plates as well as the potential difference between the plates increase, and the charge movement from the battery terminals to and from the plates decreases. All charge movement stops when the potential difference between the plates becomes equal to the potential difference between the battery terminals. 6
apacitors and apacitance Units: 1 farad = 1 F 1 /V 7
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apacitance of Parallel Plate apacitor Initial: V = 0, V = V = V. 1 w1 w2 2 bat Final: V = V = Ed, V = V = 0. bat w1 w2 E = Q ε A Units: 1 farad = 1 F 1 /V 0 Q = V 9
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ombining apacitors in Parallel Parallel: Same V, but different Qs. equivalent parallel Q Q + Q + Q + total = = V 1 2 3 1 2 3 = + + + 1 2 3 V Q Q Q V V V = + + + 11
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Another Way to Look at apacitors in Parallel Take a parallel plate capacitor, and divide it into two parts (transversely) The plate separation and the potential difference haven t changed The charge and the area still add up q1 + q2 q1 q2 tot = = + = 1 + V V V 2 d d q 1 A A 1 A 2 q q q 2 V V For capacitors in parallel the capacitances simply add q 1 q 2 13
ombining apacitors in Series Series: Same Q, but different Vs. series Q Q = = V V + V + V + 1 2 3 Q = V + V + V + 1 2 3 ( V / Q) ( V / Q) ( V / Q) 1 2 3 1 2 3 1 Q 1 Q 1 = + + + 1 = 1/ + 1/ + 1/ + 14
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Another Way to Look at apacitors in Series Take a parallel plate capacitor, and divide it into two parts (longitudinally) 1 tot Imagine a piece of foil in the gap, then expand the foil This time the area and charge haven t changed The plate separation and potential difference add up = V 1 + V q 2 V1 = q V + q 2 = 1 1 1 + 2 d d 1 d 2 d 1 d 2 A A A q q q q q q q q q q V V 1 V 2 V 1 V 2 16
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Example: A apacitor ircuit Find the charge and potential difference across each capacitor shown in the figure. 19
Example: A apacitor ircuit Find the charge and potential difference across each capacitor shown in the figure. 20
Example: A apacitor ircuit Find the charge and potential difference across each capacitor shown in the figure. Q V = + parallel 1 2 1 1 1 = + series 1 2 21
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If V 0 = 10 V and 1 = 4 F, find the charge stored on 1 when the switch is thrown to the left (assume that it becomes fully charged). 24
After 1 is fully charged, the switch is then thrown to the right. If 2 is 3 F and 3 is 6 F, find the final charge and voltage on each of the capacitors. 25
After 1 is fully charged, the switch is then thrown to the right. If 2 is 3 F and 3 is 6 F, find the final charge and voltage on each of the capacitors. 26
After 1 is fully charged, the switch is then thrown to the right. If 2 is 3 F and 3 is 6 F, find the final charge and voltage on each of the capacitors. 27
apacitors I What is the equivalent capacitance, eq, of the combination below? 1) eq = 3/2 2) eq = 2/3 3) eq = 3 4) eq = 1/3 5) eq = 1/2 o eq o
apacitors I What is the equivalent capacitance, eq, of the combination below? 1) eq = 3/2 2) eq = 2/3 3) eq = 3 4) eq = 1/3 5) eq = 1/2 The 2 equal capacitors in series add up as inverses, giving 1/2. These are parallel to the first one, which add up directly. Thus, the total equivalent capacitance is 3/2. o eq o
apacitors II How does the voltage V 1 across the first capacitor (( 1 ) compare to the voltage V 2 across the second capacitor (( 2 )? 1) V 1 = V 2 2) V 1 > V 2 3) V 1 < V 2 4) all voltages are zero 2 = 1.0 µf 10 V 1 = 1.0 µf 3 = 1.0 µf
apacitors II How does the voltage V 1 across the first capacitor (( 1 ) compare to the voltage V 2 across the second capacitor (( 2 )? 1) V 1 = V 2 2) V 1 > V 2 3) V 1 < V 2 4) all voltages are zero The voltage across 1 is 10 V. The combined capacitors 2 + 3 are parallel to 1. The voltage across 2 + 3 is also 10 V. Since 2 and 3 are in series, their voltages add. Thus the voltage across 2 and 3 each has to be 5 V, which is less than V 1. 10 V 2 = 1.0 µf 1 = 1.0 µf 3 = 1.0 µf
apacitors III How does the charge Q 1 on the first capacitor (( 1 ) compare to the charge Q 2 on the second capacitor (( 2 )? 1) Q 1 = Q 2 2) Q 1 > Q 2 3) Q 1 < Q 2 4) all charges are zero 2 = 1.0 µf 10 V 1 = 1.0 µf 3 = 1.0 µf
apacitors III How does the charge Q 1 on the first capacitor (( 1 ) compare to the charge Q 2 on the second capacitor (( 2 )? 1) Q 1 = Q 2 2) Q 1 > Q 2 3) Q 1 < Q 2 4) all charges are zero We already know that the voltage across 1 is 10 V and the voltage across both 2 and 3 is 5 V each. Since Q = V and is the same for all the capacitors, then since V 1 > V therefore Q 2 1 > Q. 2 10 V 2 = 1.0 µf 1 = 1.0 µf 3 = 1.0 µf