PHYSICS 8B, FALL 2014Lecture 2, Midterm 1

C. Bordel

Thursday, October 9th, 7pm-9pm

Make sure you show all your work and

justify your answers in order to get full credit.

Problem 1 – Charged rod (25pts)

L

Q

P

d

x

y

A uniformly charged straight wire of negligiblethickness has length L and carries charge Q > 0.Determine the direction and magnitude of theelectric field created by the finite-size wire at pointP which is a distance d from the wire along theperpendicular bisector. Hint: Integral formulasare provided in the equation sheet for you to eval-uate the definite integral.

Problem 2 – Spherical capacitor (25pts)

+Q

R1

R2

�Q

A capacitor is made of two concentric spherical shellsof radii R1 and R2, carrying the charges +Q and −Qrespectively, where R1 < R2.

(a) Explain which method you are going to use to calcu-late the electric field produced by a uniform spher-ical charge distribution.

(b) Determine the electric field inside and outside thecapacitor.

(c) Express the voltage across the capacitor.

(d) Determine the capacitance C of the sphericalcapacitor.

1

Problem 3 – Current, resistivity and power (25pts)

r1r2

L

⇢

Figure 1

D0

L0

2D0

2L0

Figure 2

A hollow cylindrical resistor is made of a conducting materialof resistivity ρ, length L, inner radius r1 and outer radius r2 asshown in Figure 1.

(a) Assuming that the current passes along the direction ofthe symmetry axis of the resistor, calculate the resistanceR.

(b) The same voltage V is applied at the two ends of twosolid thick conducting cylinders (see Figure 2) made ofthe same material of resistivity ρ such that the currentsflow along the same direction as the symmetry axis. Ifone conductor is twice as long and twice the diameter ofthe second, what is the ratio of the current through thefirst relative to the second?

(c) The wiring of a house must be thick enough to preventany fire hazard. What should the diameter D of a solidcylindrical wire be if it is to carry a maximum current Iand produce power losses of no more than P per meterof length?

(d) Recalculate the resistance R of the hollow cylindricalresistor pictured in Figure 1, now assuming that thecurrent is radial .

Problem 4 – RC Circuit (Conceptual questions) (25pts)

C

R

The capacitor is initially uncharged.

(a) If this is connected to a battery that sources a voltage E ,what is the maximal charge that can be reached on thecapacitors plates? You don’t need to do any calculation;the solution of the differential equation is provided on theequation sheet.

(b) How would you change the resistance R to double the time required to reach a given fractionof the maximum charge? Explain.

(c) Determine, with a minimum of calculation, the current going through the equivalent circuitin the following two cases:i) immediately after the battery is connected to the circuit.ii) a long time after the battery is connected to the circuit.

(d) Redraw the circuit using a replacement for the capacitor in the short-term and long-termlimits. Your effective circuits should capture the current and voltage in the short–term andlong–term limits.

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Midterm 1 Equation Sheet

• ~F =1

4πε0

Q1Q2

r2r̂

= kQ1Q2

r2r̂

• ~F = Q~E

• ~E =1

4πε0

∫dQ

r2r̂

= k

∫dQ

r2r̂

• λ =dQ

d`

• σ =dQ

dA

• ρ =dQ

dV

• ΦE =

∫A~E · d ~A

•∮A~E · d ~A =

Qencε0

• ~p = Q~d

• ~τ = ~p× ~E

• U = −~p · ~E

• ∆U = Q∆V

• dV = − ~E · d~̀

• V =1

4πε0

∫dQ

r

= k

∫dQ

r

• Q = CV

• Ceq = C1+C2 (in parallel)

• 1

Ceq=

1

C1+

1

C2(in series)

• ε = κε0

• U =Q2

2C

• I =dQ

dt

• V = IR

• R = ρ`

A

• P = IV

• I =

∫A~j · d ~A

• ~j = nq~vd =~E

ρ

• Req = R1 +R2

(in series)

• 1

Req=

1

R1+

1

R2

(in parallel)

•∑

juntion

I = 0

•∑loop

V = 0

• Q(t) = CE(

1− e−t/(RC))

(RC Circuit, charging)

• Q(t) = CEe−t/(RC)

(RC Circuit, discharging)

•∫xm dx =

xm+1

m+ 1for m 6= −1

•∫

1

xdx = lnx

In the following, a is a constant:

•∫

dx√x2 + a2

= ln(x+

√x2 + a2

)•∫

1

x2 + a2dx =

1

atan−1

(xa

)•∫

1(x2 + a2

)3/2 dx =x

a2√x2 + a2

•∫

x(x2 + a2

)3/2 dx = − 1√x2 + a2

•∫

sinx dx = − cosx

•∫

cosx dx = sinx

• cos 0 = − cosπ = 1

• sin(π2

)= − sin

(3π2

)= 1

• cos(π2

)= sin 0 = sinπ = 0

3