Description
PHYS 2401 WEEK 5
EX. without numbers
Find the total potential energy(Ue).
The four charges that make up a square are equivalent. All side lengths are ‘d’ Items known: m, Q, and d
For this problem, we can use Ue = (1/2)*Q*(V) where
ΔV = E*d
E = ke(Q/d2)
So
Ue = (1/2)*Q* ke*(Q/d2)*d = (1/2)*Q2*ke/d
Utotal = U12 + U13 + U14……. ???? This is the long way
However, since we know all of the charges are the same we can do it a simpler way: Utotal = 8((1/2)*Q2*ke/d) + 4((1/2)*Q2*ke/(sqrt(2)*d)) = (4 + sqrt(2))*ke* Q2/d
Note: We can only do this because they are the same charge
In the above equation, “8((1/2)*Q2*ke/d) “ is derived for the linear paths of energy (think, tracing the box). Each charge has two other charges acting on it in a linear path so 2*4=8. The second part “4((1/2)*Q2*ke/(sqrt(2)*d))” is derived for the diagonal paths of energy (cutting across the box). Each charge has a single charge acting on it through a diagonal path so 1*4=4. The denominator is “ sqrt(2)*d” because of the 45-45-90 special triangle.
Don't forget about the age old question of Which is more important vision or hearing?
We also discuss several other topics like How do changes in population affect politics?
Okay now let Q in the lower right corner fly away. What is its speed when ∞ far away?
First lets look at a similar but simpler case.
Initial Case:
Final Case: r2=∞
Utotal = (k*Q1*Q2/r1)+ (k*Q1*Q2/r2) ???? (k*Q1*Q2/r2) = 1/∞ = 0 Remember: ΔK = - ΔUe
= (1/2)mv12 -(1/2)mv22 ???? (1/2)mv12 cancels out because v1 = 0
So
(k*Q1*Q2/r1) = (1/2)mv22
Now Back to the harder question
Utotal = Uinital = (4 + sqrt(2))*ke* Q2/d Don't forget about the age old question of What is the definition of sexual behavior?
Ufinal = (2*ke* Q2/d) + (ke* Q2/(sqrt(2)d)) ???? To find this, redue the steps that found Uintial but now theres only 3 charges
Again,
ΔK = - ΔUe
= (1/2)mv12 -(1/2)mv22 ???? (1/2)mv12 cancels out because v1 = 0 So,
(1/2)mv22 = (ke* Q2/d)((4 + sqrt(2)) – (2 + 1/sqrt(2))) If you want to learn more check out What are the challenges to feature theories?
To find v2, just rearrange the above equation
Equipotential Surfaces
ΔV = - ∫ E * ds
Ex = -dV/dx
Ey = -dV/dy
Ez = -dV/dz
Potential of a Conductor
V is constant throughout the conductor
ΔV = - ∫ E * ds ???? V is CONSTANT because E = 0, this DOES NOT mean V = 0
EX. without numbers
Voutside = k*Qtotal/R
Vinside = k*Qtotal/r ???? This is constant
EX. Where is is the potential difference equal to zero? V = 0 ? Q1 = -20uC ; Q2 = 10uC
V = k*Q1/r
Vtotal = V1 + V2
Vinside(x) = 0 = (k*(-20uC)/x)+( k*(10uC)/(1m – x))
(k*(20uC)/x) = ( k*(10uC)/(1m – x)) ????’k’ and the units ‘uC’ cancel out (2/x) = (1/(1-x)) Don't forget about the age old question of What is ethnomethodology perspective?
2 – 2x = x
2 = 3x
X = 2/3 meters
Voutside(x) = 0 = (k*(10uC)/(x-1m))+( k*(-20uC)/(x))
(k*(10uC)/(x-1m) = ( k*(-20uC)/(x)) ????’k’ and the units ‘uC’ cancel out 1/(x-1) = (2/x)
2x - 2 = x
X = 2 meters
EX. Find Q1,Q2
r1 = 4cm ; r2 = 6cm ; Qtotal = 20uC
Vtotal = V1 + V2
V1 = V2
V = k*Q1/r
k*Q1/r1 = k*Q2/r2 ???? ‘k’ cancels out
Q1/r1 = Q2/r2
Q1/4cm = Q2/6cm
Q2 = 1.5*Q1
Qtotal = Q1 +Q2
20uC = Q1 + 1.5Q1 = 2.5Q1
Q1 = 8uC
Q2 = 12uC
Chapter 26 Capacitors
A capacitor is made up of two conductors.
When the capacitor is charged, the conductors carry charges of equal magnitude but opposite sign We also discuss several other topics like What is saturated hydraulic conductivity?
ΔV = - ∫C E*ds
Q = E*C ???? ‘C’ stands for capacitance. Units: Farads(F)
1 F = 1 C/V ???? farad to coulombs/volts
Parallel Plate Capacitors
How do we find C of pair of conductors?
1) Assume some value of charge
2) Find E between conductors
3) Find V.
4) C = Q/V
E = σ/
σ = Q/A
So,
E = Q/(A* ) ???? first equation for E
V= E*d
E = V/d ????second equation for E
So,
Q/(A* ) = V/d
Q = ((A* )/d)*V ???? remember Q =C*V
So,
C = ((A* )/d) ???? parallel plate capacitor capacitance
Cylindrical Capacitor
Side View Ariel View
λ = Q/ l
E = 2*k* λ /r
V = ∫ba 2*k* λ /r *dr
= 2*k* λ* ln(r) |ba
= 2*k* (Q/l) * (lnb – lna)
Note : 2k = 1/(2*pi* )
= (1//(2*pi* )) *(Q/l)*ln(b/a)
Q = (2*pi* )/(ln(b/a)) V
So,
C = (2*pi* )/(ln(b/a)) ???? Cylindrical capacitor capacitance Spherical Capacitor
C = 4*pi* *a*b/(b-a)
Capacitor hooked up to a battery
V stays constant
d‘ = 2d ???? ‘d’ is the distance between the plates of a parallel plate capacitor Q’ = C’V
Q’ = Q/2
C’ = *A/d’ = (1/2)C
What happens when the battery is disconnected?
Then d’ = 2d
Q stays the same
C’ = C/2
V= Q/C
V’ = 2V
V’ = Q/C’
Symbols
Capacitor -| |-
Battery
Switch
Capacitors in Parallel
To combined C1 and C2 to find a single capacitor equal to C1 and C2 when they are in parallel Ceq = C1 +C2
Because,
Vb = V1 = V2
Qb = Q1 +Q2
Qb = C1*Vb + C2*Vb = (C1 + C2)Vb
Qb/Vb = (C1 + C2)
Ceq = C1 +C2
Capacitors in Series
To combined C1 and C2 to find a single capacitor equal to C1 and C2 when they are in series 1/Ceq = 1/C1 +1/C2
Because,
Qb = Q1 = Q2
Vb = V1 +V2
Vb = Qb/C1 + Qb/C2 = (1/C1 +1/C2)Qb
Vb/Qb = (1/C1 + 1/C2)
1/Ceq = 1/C1 +1/C2
Can also be written as :
Ceq = 1/(1/C1 +1/C2) = C1*C2/(C1+C2)
EX.
5uF + 4uF = 9uF
2uF + 3uF + 7uF = 12uF
1/(1/9uF + 1/3uF) = 9/4 uF 1/(1/6uF +1/12uF) = 4uF
9/4uF + 4uF = 25/4uF
Qb = Ceq*Vb = (25/4 uF)*(10V) = 25/40 uC EX.
6uF +3uF =9uF
1/(1/9uF +1/10uF) = 4.74 uF
Qb = Ceq*Vb = (4.74uF)(10V) = 47.4uC
Remember capacitors in series have the same Q So, Qb = Q10uF = Q9uF
Now to find Q6uF and Q3uF from the original circuit, but first find the voltage for capacitor 9uF because we know capacitors in parallel have equal value
10 V – V10uF – V9uF =0
V9uF = 10V - V10uF
= 10V -Q10uF/ C10uF
= 10V – 47.4uC/10uF =5.25 V = V6uF = V3uF
So, Q = CV
Q6uF = C6uF* V6uF = (6uF)*(5.26V) = 31.6uC
Q3uF = C3uF* V3uF = (3uF)*(5.26V) = 15.8uC
If you increase C10uF, will Q6uF and Q3uF stay the same , increase, or decrease? C10uF increases, Ceq decreases
Qb decreases, V10uF decreases
so,
V6uF, V3uF both increase
Q6uF, Q3uF both increase
EX.
C1 = 15uF
C2 = 6.67uF
1) Close S1, leave S2 open
Q1 = C1*Vb = 15uF *12V = 180uC
Q2 = 0
2) Open S1(battery out of play), close S2. What are the final values of Q’1,Q’2,V’1,and V’2?
Q’1 + Q’2 = Q1
Q’1 + Q’2 = 180uF
V’1 = Q’1/C1 +Q’2/C2 = V’2
Q’2 =(C2/C1)Q’1 = .444Q’1
Q’1 +.444Q’1 =180uF
1.444Q’1 = 180uF
Q’1 =125uC
Q’2 = 55uC
V’1 = Q’1/C1 = 125uC/15uF = 8.3V =V’2
EX. Find C1 and C2 when in series the equal 5uF, and in parallel they equal 30uF C1*C2/(C1+C2)=5uF
C1 +C2 =30uF ???? C2 = 30uF – C1
Plug in equation 2 into equation 1
C1(30uF – C1)/(C1+30uF-C1) = 5uF
-1/30(C1)2 + C1 -5 =0
Solve the quadratic equation to get Q1 then solver for Q2
The answers are: Q1=23.65; Q2 = 6.35
Battery Work
-stored energy in C
Vi = 0
dWb = Vc*dq
Vc = q/C
dWb = (q/C)*dq Wb = U
Wb = (1/2)Q2/C = (1/2)CVb2 = (1/2)QVb
EX. Energy stored? C = 20uF, Vmax = 109 U = (1/2)(2*10-5uF)(109V)2= 1013J