Electric Potential Energy of a Charge (continued) i is “the” reference point. Choice of reference point (or point of zero potential energy) is arbitrary. i is often chosen to be infinitely far ( )

Scalar! Electric Potential U(r) of a test charge q0 in electric field generated by other source charges is proportional to q0 . So U(r)/q0 is independent of q0, allowing us to introduce electric potential V independent of q0. taking the same reference point [Electric potential] = [energy]/[charge] SI units: J/C = V (volts) A positively charged particle produces a positive electric potential; a negatively charged particle produces a negative electric potential. Scalar! Potential energy difference when 1 C of charge is moved between points of potential difference 1 V

Two Ways to Calculate Potential Integrate -E from the reference point (∞) to the point (P) of observation: Q P r A line integral (which could be tricky to do) If E is known and simple and a simple path can be used, it may be reduced to a simple, ordinary 1D integral. Integrate dV (contribution to V(r) from each infinitesimal source charge dq) over all source charges: q1 q2 q3 q4 P P Q

Equipotential Surfaces An equipotential surface is a surface on which the potential is the same everywhere. For a displacement Δr of a test charge q0 along an equipotentital, U = constant on an equipotential surface. E is everywhere perpendictular to U Potential difference between nearby equipotentials is approximately equal to E times the separation distance. Equipotential surfaces are drawn at constant intervals of ΔV Moving along an equipotential surface keeps the potential constant, so the electric field can’t change on the surface... therefore the surface must be perpendicular. http://www.its.caltech.edu/~phys1/java/phys1/EField/EField.html

Potential and Conductors Equipotential surfaces Entire conductor including its surface(s) has uniform V. Draw equipotential surfaces outside the conductor on which V is uniform. For Δl on equipotential, or equipotential surface

Electrostatic Potential Energy of Conductors bring dq in from ∞ q+dq,v+dv Spherical conductor q,v R q v dq Q V Total work to build charge from 0 up to Q:

Capacitance The two conductors hold charge +Q and –Q, respectively. Capacitor plates hold charge Q The two conductors hold charge +Q and –Q, respectively. The capacitance C of a capacitor is a measure of how much charge Q it can store for a given potential difference ΔV between the plates. Expect Capacitance is an intrinsic property of the capacitor. farad (Often we use V to mean ΔV.)

Energy of a charged capacitor How much energy is stored in a charged capacitor? Calculate the work required (usually provided by a battery) to charge a capacitor to Q Calculate incremental work dW needed to move charge dq from negative plate to the positive plate at voltage V. Total work is then

Steps to calculate capacitance C Put charges Q and -Q on the two plates, respectively. Calculate the electric field E between the plates due to the charges Q and -Q, e.g., by using Gauss’s law. Calculate the potential difference V between the plates due to the electric field E by Calculate the capacitance of the capacitor by dividing the charge by the potential difference, i.e., C = Q/V.

Example: (Ideal) Parallel-Plate Capacitor Put charges +q and –q on plates of area A and separation d. Calculate E by Gauss’s Law b 3. Calculate V by a 4. Divide q by V q is indeed prop. to V C is prop. to A C is inversely prop. to d

Ideal vs Real Parallel-Plate Capacitors Uniform E between plates No fields outside Often described as “thin” Non-uniform E, particularly at the edges Fields leak outside

READING QUIZ 1 Consider a parallel plate capacitor with C = 1 μ F is connected to a battery which has voltage V = 40 V. Disconnect the battery. What is the energy stored in the capacitor ? The plate separation is changed from d to d/3. Is the energy stored in the capacitor reduced by 2/3 or tripled ? 1.6 x 10-3 J tripled 8.0 x 10-4 J reduced by 2/3 C. 6.0 x 10-3 J reduced by 2/3 8.0 x 10-4 J tripled E. 1.6 x 10-3 J reduced by 2/3

Capacitors Q A capacitor is a device that is capable of storing electric charges and thus electric potential energy. => charging Its purpose is to release them later in a controlled way. => discharging Capacitors are used in vast majority of electrical and electronic devices. +Q -Q Typically made of two conductors and, when charged, each holds equal and opposite charges.

DOCCAM 2 Capacitor Types

Physics 241 –warm-up quiz Two identical ideal parallel plate capacitors. Both initially carry charge Q. One is always connected with an ideal battery while the other is not. After increase the distance between the plates by a factor of two, which of the following statements is true. A B E field is not changed in both A and B . E field in A is half of the original value; E field in B is not changed. E field in A is not changed; E field in B is half of the original value. E field is half of the original value in both A and B . One need to know the voltage of the battery to give the answer

Numerical magnitudes Let’s say: Area A= 1 cm² , separation d=1 mm Then This is on the order of 1 pF (pico farad = 10-12 F) Generally the values of typical capacitors are more conveniently measured in µF or pF.

Long Cylindrical Capacitor Put charges +q on inner cylinder of radius a, -q on outer cylindrical shell of inner radius b. Calculate E by Gauss’ Law a +q b -q L 3. Calculate V C dep. log. on a, b 4. Divide q by V

Spherical Capacitor Put charges +q on inner sphere of radius a, -q on outer shell of inner radius b. Calculate E by Gauss’s Law 3. Calculate V from E q is proportional to V C only depends on a,b 4. Divide q by V (isolated sphere)

DOCCAM 2 Energy storage in capacitors

A + + + + d - - - - - Q: U: C: V: E: By how much? Example Suppose the capacitor shown here is charged to Q and then the battery disconnected. Now the plates are pulled apart so that the final separation is d1 (> d ). How do the quantities Q, U, C, V, E change? What if battery remains connected? Q: U: C: V: E: remains the same.. no way for charge to leave. increases.. add energy to system by separating decreases.. since d increases increases.. since C decreases, but Q remains same remains the same... depends only on charge density By how much?

Where is the energy stored? Energy is stored in the electric field itself. Think of the energy needed to charge the capacitor as being the energy needed to create the electric field. To calculate the energy density in the field, first consider the constant field generated by a parallel plate capacitor, where ++++++++ +++++++ - - - - - - - - - - - - - - - Q +Q This is the energy density, u, of the electric field…. The electric field is given by: Þ The result applies to general cases. The energy density u in the field is given by: Units:

Physics 241 – 9:30 Quiz 3 February 3, 2011 If we are to make an ideal parallel plate capacitor of capacitance C = 1 μF by using square plates with a spacing of 1 mm, what would the edge length of the plates be? Pick the closest value. 10 m 1 cm 100 m 1 mm 1 m

Physics 241 –10:30 Quiz 3 February 3, 2011 We have an ideal parallel plate capacitor of capacitance C=1 nF (n=10-9) that is made of two square plates with a separation of 2 μm. A battery keeps a potential difference of 15 V between the plates. What is the electric field strength E between the plates? 30 N/C 5 x 109 V/m 15,000 V/m 7.5 x 106 N/C Not enough information to tell

Physics 241 –Quiz 3 11:30 February 3, 2011 We have an ideal parallel plate capacitor of capacitance C=1 μF that is made of two square plates. We could double the capacitance by doubling the separation d doubling charges on the plates Q halving potential difference between the plates V increasing the edge length a of the plates by a factor of 1.4 halving the area of the plates A a d A a Q -Q