Lecture 14

Readings

http://umdberg.pbworks.com/w/page/63915042/The%20capacitor%20%282013%29

http://umdberg.pbworks.com/w/page/105606864/The%20dielectric%20constant

Capacitors, Continued

• Relate this to $\mathrm{\Delta }V$$\Delta V$ , difference in electric potential , between plates

• Reminder: $E$$E$ is the negative slope of $V$$V$

  • If $V$$V$ is linear, slope of $V$$V$ is just $\frac{\mathrm{\Delta }V}{d}$$\frac{\Delta V}{d}$

• Rearrange equation:

• Define quantity of 'C' , called the capacitance:

• $Q=Capacitance(C)\ast \mathrm{\Delta }V$$Q = Capacitance\ (C)* \Delta V$

  • Capacitance depends only on geometry, area, and distance
  • If you know capacitance and voltage difference, you can determine $Q$$Q$ on each plate

• Energy Stored in a Capacitor:

Dielectric Constant

• When matter is exposed to charge, we know that the charge "pulls" opposite charge in matter towards itself

• Electric field is created by charges on plates:

• Put material between plates:

• $E$$E$ due only to rearrangement of charges inside the material is to the left ( points from positive to negative )
• So $E$$E$ from charges on plates is to the right, and $E$$E$ from charges inside the material is to the left

• The net $E$$E$ inside the material is weaker (smaller) than it would be without the material

  • The material has "covered up" some of the $E$$E$ field that would have been there if there were a vacuum between the plates

• We say the material has been polarized by the $E$$E$ field.

  • The material itself now has a "left" side and a "right" side ( with negative and positive charges ).

    • It has "poles"

• How much is $E$$E$ reduced?

  • Depends on the material:

• $k$$k$ is called the "dielectric constant".

  • Always greater than or equal to 1

• Also changes capacitance.

  • $C$$C$ becomes:
  $$C=\frac{k\ast A}{4\ast \pi \ast k_c\ast d}$$
  • So Capacitance (C) increases due to material
  • For almost all real capacitors, material is present between plates

Conductors

• Metals

• Charges are very free to move in a donductor

  • Not bound to nucleus

• If you put a conductor between the plates of a capacitor, the charges exactly match the outer plates:

• So $E$$E$ from plates is to the right, and $E$$E$ due to charges on the conductor is to the left
• Because charges exactly match, the net E inside the conductor = 0
• In fact, in all charge situations (with nothing moving) , the electric field inside a conductor is zero
• So if $E$$E$ exists inside a conductor, free charges will feel force and move

Clicker Questions