High Electric Field at Sharp Tips

1. Potential is the same: The smaller the radius of curvature, the larger the electric field. High Electric Field at Sharp Tips Two conducting spheres are connected by a long conducting wire. The total charge on them is Q = Q1+Q2. With same potential, sphere with smaller radius carry smaller amount of charge

2. Equipotential surfaces Potential and Conductors • Entire conductor including its surface(s) has uniform V. • E  an equipotential surface everywhere • Equipotential surfaces are drawn at constant intervals of V • Potential difference between nearby equipotentials is approximately equal to E times the separation distance.

3. q1 r13 r12 q2 r23 q3 Again, Electrostatic Potential Energy The electrostatic potential energy of a system (relative to∞)is the (external) work needed to bring the charges from an infinite separation to their final positions. Generally for n charges:

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

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

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

7. Put chargesQ and -Q on the two conductors, respectively. Calculate the electric field Ebetween the plates due to the charges Q and -Q, e.g., by using Gauss’s law. Calculate the potential difference Vbetween 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. Steps to calculate capacitance C

8. 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

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

10. Physics 241 –warm-up quiz Two identical ideal parallel plate capacitor. 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

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

12. +q • C dep. log. on a, b 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 b -q L 3. Calculate V 4. Divide q by V

13. 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)

14. Calculate incremental work dW needed to move charge dq from negative plate to the positive plate at voltage V. Total work is then 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

15. - Q • To calculate the energy density in the field, first consider the constant field generated by a parallel plate capacitor, where - - - - - - - - - - - - - - ++++++++ +++++++ +Q This is the energy density, u, of the electric field…. Þ Units: 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. • The electric field is given by: • The energy density u in the field is given by:

16. Physics 241 –Quiz 6a If we are to make an ideal parallel plate capacitor of capacitance C = 1 F 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

17. Physics 241 –Quiz 6b We have an ideal parallel plate capacitor of capacitance C=1 nF that is made of two square plates with a separation of 2 m. 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

18. a Q A d Q Physics 241 –Quiz 6c We have am ideal parallel plate capacitor of capacitance C=1 F 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