Fields in Parallel Plates- Capacitors Physics Mrs. Coyle

Electric Fields Between Parallel Plates Uniform Electric Field of Strength: E = V / d d is the distance between the plates Units: V/m =N/C=J/(Cm)

Force Experienced by a Test Charge Inside the Parallel Plate Field The field is uniform so at any point in the field a test charge would feel a force F = qE

Work and Energy V=Ed V=W/q 0 W/q 0 = Ed W=q 0 Ed Unit for work: Joule or the electron volt 1 eV= q e 1V= 1.6 x C x 1V 1eV= 1.6 x J

Problem 1 Find: a)The field inside the plates. b)The force felt by a 9nC charge that is situated in the field.

Continued c)Draw the equipotential surfaces between the plates (hint: equally spaced and parallel) d)If there three equipotential surfaces,what would be the distance between each? e)The work to move the charge from one equipotential to the other.

Continued f)The potential difference between each equipotential surface and the negative plate.

Equipotentials and Field Lines

Equipotentials and Field Lines

Note Zero work is done when a charge is moved along an equipotential surface.

Milikan’s Oil Drop Experiment He found the charge of the electron e = x C by noting that the droplets always carried whole number multiples of that number.

Capacitor

A capacitor stores charge (stores energy).

Capacitors

Leyden Jar 1746, Dutch physicist Pieter Van Musschenbroek Lived in “Leyden” Used by Ben Franklin

Capacitance Capacitance C = Q/V Unit: Farad (F)= coulomb/ volt Q charge stored in capacitor V voltage across the capacitor

Problem What is the charge of a 2µC capacitor that is charged, so that the potential difference is 8V?

Question If the charge of a capacitor is changed, does that affect the potential difference or the capacitance?