Chapter 19 - Electric Potential Flashcards
How to represent the work from a to b in terms of potential energy?
Watob = mgha - mghb
= UA-Ub
= -∆U
Determine the number of particles, each carrying a charge of 1.60×10−19C
(the magnitude of the charge on an electron), that pass between the terminals of a 12-V
car battery when a 60.0-W headlight burns for one hour
Find the energy
Energy = power x tiem
2.2 x 10^5
Use V = U/q to find charge q
q = Ne
= 1.1 x 10^23
Distinguish between electric potential (V) and electrostatic potential energy (EPE/U)
V = kq/r
Property of electric field, how much energy unit charge has at this location
Dufferece in EPE per unit charge
U = kq1q2/r
Property of system, total energy in system
EPE is essentially the system version of V
Energy to move charged particle in an electric field
A particle has a mass of 1.8x10-5kg and a
charge of +3.0x10-5C. It is released from
point A and accelerates horizontally until it
reaches point B. The only force acting on
the particle is the electric force, and the
electric potential at A is 25V greater than
at B.
(a) What is the speed of the particle at
point B?
Use the equation
1/2mvb^2 + Ub = 1/2mv^2A + Ua
inital velocity is sassumed to be 0 and hten you have Ua - Ub which is the same as q (Va-Vb) and Va-Vb is 25V
So
1/2mvB^2 = q(vA-vB)
input and solve for Vb and you should get 9.1
A point charge Q = +1.20 µC is held fixed at the
origin. A second point charge q = +4.60 µC with mass
of 2.80 × 10–4 kg is placed on the x axis, 0.250 m from
the origin. (a) What is the electric potential energy U
of the pair of charges? (b) The second point
charge is released from rest. What is its speed when its
distance from the origin is (i) 0.500 m
a)
Simply do U = kqQ/r
you should get 0.198J
b) Initial U is what we just had, final U is what we get when we set r = 0.5 and this is 0.999J
Input variables
Ua = Ub + 1/2mv^2
0.198 = 0.999 + 1/2mv^2
isolate and solve for vB and you should get 26.6
Capacitance Definition
And dielectric
Ability of system to store electric charge
With greater dielectric constant, the system can insulate to a greater extent so hold charge better
Dielectric increases capacitance but decreases the field. Opposes original field, so less voltage needed to store same charge so capacitance increases
How does changing the distance between parallel plates in a capacitor affect the electric field?
Has no effect becuase both Q and A remain constant
In the television picture tube, electrons strike the
screen after being accelerated from rest through a
potential difference of 25 000 V. The speeds of the
electrons are quite large, and for accurate calculations
of the speeds, the effects of special relativity must be
taken into account. Ignoring such effects, find the
electron speed just before the electron strikes the
screen
W = qV
Work is converted into kinetic energy (ignoring relativistic effects)
1/2mv^2 = qV
solving for v
v = sqrt(2qV/m)
you get v = 9.37 x10^7
EXAM PROBLEM
Two point charges: q1 = – 4.80x10–6 C is at the origin of coordinate axes and q2 = + 3.25x10–6 C is
15.0 cm apart and on the y-axis (as shown in Figure). Point A is 8.0 cm away from the origin on the x-axis;
point B is at x = 8.0 cm and y = 6.0 cm. A charge q0 = – 2.00x10–6 C with mass m = 10– 8 kg starts moving from
stationary at point A and travels to point B. Find:
[5] (2.a) The work done by the electric field on the charge q0.
[5] (2.b) The speed of q0 at point B
PART A
a) Find potential difference at point A by adding the difference from q1 and q2. Use trigonometry to figure out the distances between points
Va = -3.71 x 10^5
Repeat the same with point B and you should get
5.61×10^4
Use the equation -Wab = q0(Va-Vb)
Wab = -(-2x10^-6)*(5.61x10^4-(-3.71x10^5)
= 0.854J
Work is equal to the change in kinetic energy, since the object starts from stationary
W = 1/2mv^2
vb = 1.31 x 10^4 m/s
How can you find the direction of the electric field?
Like if negative charge was moving downwards
Field lines always point from positive to negative charges
Thus the velocity of a positive charge moves with the electric field. Velocity of negative charge moves against electric field, so in the example above, the field would point upwards
If an electron of
mass m and charge 2e is accelerated from rest through an accelerating potential V, show that the speed it
gains is v = sqrt(4eV/m).
Use Ka + Ua = Kb + Ub with Ka = 0
Va - Vb = V
Kb = Ua-Ub
Kb = q(Va-Vb)
1/2mv^2 = q(V)
Rearrange and isolate
or work = vq
vq = 1/2mv^2
Fill in the blanks
pos charge goes from ____ electric potential to ____ electric potential
neg charge goes from ____ electric potential to ____ electric potential
1) higher to lower
2) lower to higher
this is because electric field goes from positive to negative, so positive charge moves with field
for negative charge moves opposite to electric field
Positive and negative charges and work moving with and agaisnt the field
POS CHARGE
W/ Field - Pos work
Agaisnt Field - Neg work
NEG CHARGE
W/ Field - Neg work
Agaisnt Field - Pos Work
In the circuit shown in Figure, the potential difference across ab is + 24.0 V. Calculate (a) the charge on
each capacitor and (b) the potential difference across each capacitor.
1) Calculate equivalent capacitance 3.43
2) Q = CV
8.23 x 10^-5
3) In capacitors, charge is shared in series
So V3 and V12
V3 = 8.23/6 = 13.7
V12 = 8..23/8 = 10.3
4) Now you have voltage on each capacitor, you can find charge in each.
Q1 10.3 x 3 = 3.09
Q2 10.3 x 5 = 5.15
Q3 = 8.23
