PHYS 2401 WEEK 5  

What determines the charge of an atom?

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  

How do you rearrange an equation?

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.  

What is v in electrostatics?

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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