Quiz 2 - Chapter 19 Studying Flashcards
Two equipotential surfaces surround a +1.50 × 10−8 C point charge. How far is the 190-V surface from the 75.0-V surface?
Find the potential difference for each of the two surfaces, solving for r
Imagine a radial circle around the point charge (like the washer method in calc), the distance between the surfaces is the outer surface radius minus the inner surface radius
What are the three equations for energy in a capacitor?
E = 1/2qv
E = 1/2CV^2
E = q^2/2C
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
equilaterial triangle with sides 0.5m. three point charges (q1 = 5, q2 = 6 and q3=-2) what is epe of the system
Simple! Use
U = kq1q2/r + kq1q3/r + kq2q3/r
If the particle moves from point a to b, what is the equation for the potential difference?
V = Vb - Va
It’s always final minus inital
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
How does a positive charge and negative charge gain kinetic energy?
Positive charge when it moves from high to low potential
Just like object gains kinetic energy when falling from a height
Negative charge when it moves from low to high potential
Capacitance Definition
Ability of system to store electric charge
With greater dielectric constant, the system can insulate to a greater extent so hold charge better
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
Energy = power x time
60W x 3600s = 2.2x10^5
V = U/q
q = U/V
q = 2.2 x 10^5/12V
= 1.8 x 10^4C
Number of particles = 1.8 x 10^4/1.60x10^-19
= 1.1x10^23
Using a zero-reference potential at
infinity, determine the amount by
which a point charge of 4.0x10−8C
(stationary) alters the electric potential
at a spot 1.2 m away
Use V = kq/r
Simple :)
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
As the distance between plates in a parallel plate capacitor increases, what happens to the electric field and electric potential energy
c) E = V/d both voltage and distance increase so E increases
d) Electric potential energy increases. Voltage increases and capacitance decreases, but since voltage is squared in the term E = 1/2CV^2, it increases at a faster rate and thus casues the EPE to increase
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
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
𝛥V = -W/q
work is equal to kinetic energy. solving for work you get 𝛥KE = e𝛥V
𝛥KE also equals 1/2m(vf^2)
set these equal and isolate for vf
sqrt (2e𝛥V/m)
you should get 9.4 x 10^7
Answer with whether each of the following combinations has same/differnet charge and voltage
Resistors in series
Capacitors in series
Resistors in parallel
Capacitors in parallel
Resistors in series → Same current, different voltage
Capacitors in series → Same charge, different voltage
The reason for this makes sense and is as such:
Current = q/t so charge follows suit with the rule of current
Resistors in parallel → Same voltage, different current
Capacitors in parallel → Same voltage, different charge
In a circuit, a 3F (C1) and 5F (C2)capacitor are in parallel with each other and in series with a 6F (C3) capacitor. The potential difference between the two is 24V.
a) Calculate the charge and potential difference across each capacitor
1) Calculate the Ceq
Since 3 and 5 are in parallel their equivalent is 8F and it’s in series with 6F so the equivalent is 3.43 F
2) Calculate the total charge
Q = CV
Q = 8.23 x 10^-5
Charge is the same for circuits in series, so the C1C2 = C3 and C3 has a charge described above
Now find the voltage across each of the circuits since voltage is shared in parallel
V12 = Q12/C12
= 10.3 V
So the charge would be Q1 = V1C1 (10.3 x 3) = 3.09 x 10^-5
Q2 = (10.3)(5 x 10^-6) = 5.15 x 10^-5
With the capcitance of each it is pretty easy to find the potenital difference of each
V1 = 10.3
V2 = 10.3
V3 = 13.7
A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30
mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected
from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains
connected to the potential source so the potential difference between the plates remains constant?
a) Take the equation Energy Q^2/2C
Considering the equation for capitance C = k0A/d, if distance is halved, capcitance doubles. Back to the original equation if capacitance is doubled, then energy is halved so the new energy is 4.19J
b) Energy = CV^2/2
Same principles, if distance halved, capitance doubles, which means that energy is doubled and the new energy is 16.8J
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
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
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
A 10.0 µF parallel-plate capacitor with circular plates is connected to a 12.0 V battery. (a) What is the
charge on each plate? (b) How much charge would be on the plates if their separation were doubled while
the capacitor remained connected to the battery? (c) How much charge would be on the plates if the
capacitor were connected to the 12.0 V battery after the radius of each plate was doubled without changing
their separation?
Solve for a)
Q = CV = 1.20 x 10^-4
b) When d is doubled c is halved so Q = 60
If r is doubled, a increases by a factor of 4
pir^2 vs pi(2r)^2 = 4pir^2
C increases by factor of 4 so Q = 480
