Download Key Points Formulae

 Key Points • Moving a charged particle in an electric field generates a Formulae ∆𝑈 = −𝑞! ! 𝐸!⃗ ∙ 𝑑𝑠⃗ !
change in potential energy (U) 𝑈
𝑉 = 𝑞!
• An electric field over a given distance generates an electric potential (V) • Electric potential is directly correlated to electric field through a differential equation • A surface where all points are on the same electric potential is defined as an equipotential surface; these lines are parallel to electric field lines • There is also potential energy between any charged particles; this is also based on !
∆𝑈
∆𝑉 =
= − ! 𝐸!⃗ ∙ 𝑑𝑠⃗ 𝑞!
!
∆𝑉 = −𝐸𝑑 (uniform electric field) 𝑞!
𝑉 = 𝑘! ! 𝑟!
!
𝑞! 𝑞!
𝑈 = 𝑘!
𝑟!"
𝑑𝑉
𝐸! = − 𝑑𝑡
𝑉 = 𝑘! ∫
!"
!
(continuous distribution) the inverse square of their distance !
“Easy” Problem – 25.10 (a) Calculate the electric potential 0.250 cm from an electron. (b) What is the electric potential difference between two points that are 0.250 cm and 0.750 cm from an electron? (c) How would the answers change if the electron were replaced with a proton? .250 cm .750 cm e a) 𝑉 = 𝑘!
!
𝑞!
𝑟!
𝑚!
𝑉 = (9𝑒9 𝑁 ∙ ! )
𝐶
(−1.602𝑒 − 27 𝐶)
. 250𝑒 − 3 𝑚
!
V = -­‐5.77e-­‐14 V b) 𝑉 = (9𝑒9 𝑁 ∙
!!
)
!!
!
(!!.!"#!!!" !)
V= -­‐3.85e-­‐14 V .!"#!!! !
-­‐ (9𝑒9 𝑁 ∙
!!
!!
)
!
(!!.!"#!!!" !)
.!"#!!! !
c) If the particle were a proton, these potential differences would be positive instead of negative. “Medium” Problem – 25.15 Given two particles with 2.00-­‐mC charges and a particle with charge q = 1.28e-­‐18 C at the origin, (a) what is the net force exerted by the two 2.00-­‐mC charges on the test charge q? (b) What is the electric field at the origin due to the two 2.00-­‐mC particles? (c) What is the electric potential at the origin due to the two 2.00-­‐mC particles? a) Zero. Since the two particles on either side of the origin are of the same charge and distance from the origin, there will be no net force on the particle located at the origin. b) Zero. The net electric field at the origin will be zero for the same reasons as explained in part a. c) 𝑉 = 𝑘!
!
𝑞!
𝑟!
𝑚!
𝑉 = (9𝑒9 𝑁 ∙ ! )
𝐶
!
2𝑒 − 6 𝐶 + (2𝑒 − 6 𝐶)
. 800 𝑚
V = 45,000 V “Hard” Problem – 25.21 Two particles each with charge +2.00 μC are located on the x-­‐axis. One is at x=1.00 m, and the other is at x = -­‐1.00 m. (a) Determine the electric potential on the y axis at y=0.500 m. (b) Calculate the change in electric potential energy of the system as a third charged particle of -­‐3.00 μC is brought from infinitely far away to a position on the y-­‐axis at y=0.500 m. -­‐1.00 -­‐0.500 0 0.500 = +2e-­‐6 C a) 𝑉 = 𝑘!
!
𝑞!
𝑟!
𝑚! 2.00𝑒 − 6 𝐶 + 2.00𝑒 − 6 𝐶)
𝑉 = (9𝑒9 𝑁 ∙ ! )(
𝐶
1.00 𝑚 ! + . 500 𝑚 !
V = 32.2 kV b) 𝑈 = 𝑘!
𝑞! 𝑞!
𝑟!"
𝑚! (4𝑒 − 6 𝐶)(−3𝑒 − 6 𝐶)
𝑈 = (9𝑒9 𝑁 ∙ ! )
𝐶
1.00 𝑚 ! + . 500 𝑚 !
U = -­‐9.66e-­‐2 J = -­‐3e-­‐6 C 1.00