Andres La Rosa Portland State University Lecture Notes PH212 CAPACITANCE Parallelplates capacitor 1 2 Q Q E V 1 V 2 V 2 V 1 = 2 E E is assumed to be uniform between the plates Q Q V (Battery) V 2 V 1 = E E E A d A: Area of the plates On the other hand, we have an alternative expression for E applicable to this particular charge arrangements E = V 1 V 2 = E d Note: for the purpose of simplifying the calculations, we are going to assume that these plates are infinitely large sheets with uniformly charge density, V (Voltage across the battery) V = E d = Q = V Capacitance Capacitance
Exercise: The separation between the plates is increased, while the battery is kept connected. A d 1 E 1 Question V E 1 < E 2 E 1 = E 2 E 1 > E 2? A E 2 d 2 d 2 > d 1 V
Exercise: The battery is disconnected, and the separation between the plates is increased; Q A d 1 E 1 Q Question No battery E 1 < E 3 E 1 = E 3 E 1 > E 3? A Q d 3 Q E 3 d 3 > d 1 No battery
Exercise: The separation between the plates is increased, while the battery is kept connected. d 1 Q 1 Q 1 A E 1 Question V Q 2 Q 2 A E 2 d 2 Q 1 < Q 2 Q 1 = Q 2 Q 1 > Q 2 d 2 > d 1? V
Exercise: The battery is disconnected, and the separation between the plates is increased; A Q 1 d 1 E 1 Q 1 Question No battery Q 1 < Q 3 Q 1 = Q 3 Q 1 > Q 3? A Q 3 d 3 Q 3 E 3 d 3 > d 1 No battery
Question d A 1 E 1 A 4 V d E 1 < E 4? E 1 = E 4 E 1 > E 4 E 4 V A 4 > A 1
In general Q = C V Capacitance (Value depends on the geometry of the electrodes) For the particular case of a parallelplate capacitor Unit: Farad = F = Coulomb/Volt We use capacitors to store charge Step1 Frontal view of two parallel plates Each plate is called an "electrode" Step2 Connection to a battery Step3 After removing the batteries the plates remain charged Notice: The higher the voltage, the more charge is deposited in the electrodes. Q = CV
Capacitors also store energy The question we are addressing is: "How much energy is spent in charging a capacitor? Q Q Step1 Uncharged plates C = ε o A / d Step2 Imagine, we take an amount of charge Δq from the right plate and deposit it into the left plate. Wed repeat this process again and again. Δq Δq Step3 q q Assume, at a given time, an amount of charge q has been transferred from the right plate to the left plate. The potential difference between the plates is V = q / C Step4 ΔW ext = F ext d = ( Δq ) E d = ( Δq ) V Δq External work required to transport a small charge Δq from the right plate to the left plate ΔW ext = V Δq = ( q/c ) Δq
" The electric potential energy stored in the capacitor is considered to be stored in the electric field" d
Several capacitors connected in SERIES a V 1 V 2 V 3 Q Q Q Q Q C 1 C 2 C 3 Equivalent to Q b C equiv =? Q Q C equiv V = V 1 V 2 V 3 Q = C equiv V V = C 1 C 2 C 3 C equiv = V = ( ) C 1 C 2 C 3 1 C = 1 1 1 equiv C 1 C 2 C3 ( )
Q Q Q Q Q Q Zero net charge in this section where C 1 C 2 C 3 E E' C 1 C 2 E" C 3 Question: Is E = E'? Is E' = E"?
Several capacitors connected in PARALLEL Equivalent to On the other hand Q = V C equiv which implies, C equiv = C 1 C 2 C 3
E 3 E 2 E 1
Example C 1 C 2 = 8.95 μf What is the potential difference across the capacitors? C 1 = 3.55 μf
Extracurricular reading: Feynman Lectures on Physics, Vol. II "The ambiguity of the field energy", page 276 " all energy is source of gravitational attraction" Therefore, this theory will require that, Position of the star Light bends as it passes near the Sun Observed position of the star
Capacitors with a dielectric (no batteries connected to the plates. Hence, the charge on the plates remains constant). Experimentally it is found that the potential difference between the plates decreases. The capacitor was initially charged using a battery. Then the battery was removed K is called the dielectric constant (It is a material property)
Example What are the energies stored in the capacitor before and after introducing the dielectric? Before After U i = = U f = =
DIELECTRICS: A microscopic View Metallic plate Free charge q ( negative induced charge) q' K Neutral region q' q ( positive induced Free charge Metallic plate q q' q' q Material of dielectric constant K q q' q q'
Let's use Gauss' Law to find out the induced charge q' Gaussian surface Metallic plate q q' q' q Material of dielectric constant K = Charge inside the Gaussian surface = q q'
q' = q Gauss' Law applied to dielectric materials
Q V
What is ground? It is a very large conductor that can supply unlimited amount of charge