PHYS 155 Midterm 1 Tutorial
Topics covered: Force fields, Electric Fields, Capacitors, and Dielectrics
Notes borrowed from Eric Peach. Modified for 2016 by Thomas Truong.
Force Fields: Mover’s Perspective
• From your notes, some notational quirks:
1) Potential Energy at B with respect to A is 2J.
2) Potential Energy at B with respect to C is 4J.
• Another way to look at it
1) Energy it takes to move from A to B = Final
energy at B – Initial energy at A = 2J – 0J = 2J.
2) Energy it takes to move from C to B =
2J – (-2J) = 4J
Charge and Electric Fields
Electric Fields (E) flow from positive charge to negative.
Example 1: Energy Contours and Electric
Fields: Mover’s Perspective
• Same idea as the force fields, but
now we have to consider charge.
(a) Potential Difference, or just
Potential, from D to A (same as A
with respect to D)? Ie. What is VAD?
(b) Energy (or work required) to move
a +1C charge from A to B (same as
B with respect to A)?
(c) Energy to move a -1C charge from
A to B (same as B with respect to
A)?
Electric Field due to Point Charge in one
dimension
𝑘𝑒 𝑞
𝐸= 2
𝑟
• 𝑘𝑒 is Coulomb’s Constant, 8.99 x 109 [Nm2/C2]
• q is charge in coulombs, [C]
• E is electric field strength [N/m]
• r is distance from point charge [m]
Example 2: Principle of Superposition
Three point charges are located on a circular
arc as shown.
(a) What is the total electric field at P?
(b) What is the electric force on a -5.00nC
point charge at P?
Coulomb’s Law in one dimension
𝑘𝑒 𝑞1 𝑞2
𝐹=
𝑟2
• 𝑘𝑒 is Coulomb’s Constant, 8.99 x 109 [Nm2/C2]
• q1 and q2 are charges in coulombs, [C]
• F is electric force [N]
• r is distance from point charge [m]
• Like charges repel, opposites attract
Example 3: Electrostatic Equilibrium
Two small beads having positive charges q1 = 3q and q2 = q are fixed at
the opposite ends of a horizontal insulating rod of length d = 1:50m.
The bead with charge q1 is at the origin. As shown, a third small,
charged bead is free to slide on the rod.
At what position x is the third bead in equilibrium?
Force and Work Done by an Electric Field on a
Point Charge in 3-space
𝑭 = 𝑞𝑬
Force F on charge q due to electric field E
𝑊 = 𝑭·d
Work W done by electric field over distance d
Example 4: Work and Energy
A Particle having charge q = +2:00uC and
mass m = 0.010 kg is connected to a string
that is L = 1.50 m long and tied to the pivot
point P in the figure. The particle, string, and
pivot point all lie on a frictionless, horizontal
table. The particle is released from rest when
the string makes an angle = 600 with a
uniform electric field of magnitude E = 300
V/m. Determine the speed of the particle
when the string is parallel to the electric
field.
Parallel Plate Capacitors
|Q| =
ε0 𝐴
𝑑
|𝑉|= C |V|
• ε0 is the permittivity of free space, 8.85 x 10-12 [C2/Nm2]
• A is the area of each of the plates, [m]
• V is the voltage across the plates, [V]
• d is the distance between the plates,[m]
σ = Q/A
• σ is the surface charge density
Example 5: Parallel Plate Capacitor
When a potential difference of 150 V is applied to the plates of a
parallel-plate capacitor, the plates carry a surface charge density of
30.0 nC/cm2. What is the spacing between the plates?
Energy of a Capacitor
W=
𝑄2
2𝐶
1
= QV
2
• W is the energy of the capacitor, [J]
• Q is the charge on the capacitor in Coulombs, [C].
• Not to be confused with capacitance C, in farads [F]!
• C is the capacitance of the capacitor, [F] (not Coulombs!)
• V is the voltage across the capacitor, [V]
Example 6: Energy in a Capacitor
• A 12.0-V battery is connected to a capacitor, resulting in 54.0uC of
charge stored on the capacitor. How much energy is stored in the
capacitor?
Dielectrics and Capacitors
• Dielectrics inserted between parallel plate capacitors change the
capacitance of the capacitor.
• ε𝑟 , the relative permittivity of the dielectric, lowers the electric field between
plates.
• Lower electric field allows for more charge on the parallel plates for the same
voltage.
• More charge means a larger capacitance for the same voltage.
Dielectrics and Capacitors
𝐶𝑡𝑜𝑡𝑎𝑙 = ε𝑟 𝐶𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟
𝐸𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟
𝐸𝑑𝑖𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 =
ε𝑟
𝐸𝑡𝑜𝑡𝑎𝑙 = 𝐸𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 − 𝐸𝑑𝑖𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐
• Ccapacitor and Ecapacitor are the capacitance and electric field of the
capacitor before the dielectric is inserted.
• Ctotal and Etotal is the total capacitance and electric field after the
dielectric is inserted
• Edielectric is the electric field induced by the dielectric
Example 7: Dielectric Capacitor 1
The voltage across an air-filled parallel-plate capacitor is measured to
be 85.0 V. When a dielectric is inserted and completely fills the space
between the plates, the voltage drops to 25.0 V. What is the dielectric
constant of the inserted material?
Example 8: Dielectric Capacitor 2
A parallel-plate capacitor in air has a plate separation of 1.50 cm and a
plate area of 25.0 cm2. The plates are charged to a potential difference
of 250 V and disconnected from the source. The capacitor is then
immersed in distilled water (ε𝑟 = 80). Assume the liquid is an insulator.
Find:
(a) The charge on the plates before and after immersion
(b) The capacitance and potential difference after immersion
(c) The change in energy of the capacitor