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TTU - FIN 2401 - Class Notes - Week 5

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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.

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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

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 We also discuss several other topics like What is the meaning of sexual contact?

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))) Don't forget about the age old question of What is pattern recognition?

To find v2, just rearrange the above equation We also discuss several other topics like What is the thomas theorem in sociology?

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))

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) If you want to learn more check out How do you measure water content in soil?

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

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