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Problem 1 (25 points).
I. ANSWER (A)
We donāt know R 1 , R 2 , or R 3 , but we do know their ratios. Letās define R such that:
R 1 = R (1)
R 2 = 2R (2)
R 3 = 2R (3)
And letās call the batteryās voltage V. Then with the switch open, the current through R is
IRopen 1 = V Rtotal
= V
R 1 + R 2
= V
R + 2R
= V
3 R
When the switch is closed, Rtotal changes: the equivalent resistance of the parallel combination of R 1 and R 2 is
Requiv =
1 R 2 +^
1 R 3
1 2 R +^
1 2 R
= R (5)
So
IclosedR 1 =
V
Rtotal
V
R 1 + Requiv
V
2 R
So how much does the current change?
IRclosed 1 IRopen 1
V 2 R V 3 R
=^3
i.e., it changes by a factor of 3/2. Notice that although we wrote down the variables R and V , which we arenāt given, they donāt show up in our final answer.
II. ANSWER (B)
With the switch open,
IRopen 2 = IRopen 1 = V 3 R
With the switch closed, the current through R 1 , IRclosed 1 , is split between R 2 and R 3. Since R 2 = R 3 , the current should be split evenly:
IRclosed 2 = IRclosed 3 =^1 2
IRclosed 1 =^1 2
V
2 R
= V
4 R
So how much does the current change?
IRclosed 2 IRopen 2
V 4 R V 3 R
i.e., it changes by a factor of 3/4.
Problem 4 (25 points).
I. ANSWER
Because the two spheres are conducting and are connected by a wire, they must be at the same potential. Letās call this potential V. It will come in useful, but we wonāt end up having to calculate it. Now, for a conducting sphere of radius r at potential V , whatās the field at the surface? Letās choose V = 0 at a point infinitely far away. Letās call the charge on the sphere Q (we wonāt need to calculate this, either). Then drawing a sphereical Gaussian surface (named S) at radius R > r, we have
Q
Īµ 0
S
E Ā· dA = E(R)A = 4ĻR^2 E(R) (1)
The first equality if Gaussās law; the second equality is true because the field is spherically symmetric, so it has the same value everywhere on S, and is everywhere perpendicular to S. So
E(R) =
Q
4 ĻĪµ 0 R^2
So we can find
V (R) =
ā« R
ā
E(Rā²) Ā· dRā²^ =
Q
4 ĻĪµ 0
ā« R
ā
dRā² Rā²^2 =^
Q
4 ĻĪµ 0 R (3)
Why do we want to know V (R)? Just so that we can find E in terms of V :
E(R) =
V (R)
R
So at the surface of a conducting sphere of radius r, E = V /r. Now, returning to our small sphere, what is V? Well, itās the same as on the large sphere: V = Er 2 r 2 , so the field at the surface of the small sphere is
Er 1 = V r 1
= Er 2 r^2 r 1
The question asks for r 1 , so letās solve for it:
r 1 = r 2
Er 2 Er 1 (6)
We know that r 2 = 10 cm, Er 2 = 3 kV/cm, and air will ionize when Er 1 ā„ 30kV/cm. So
r 1 ā¤ (10cm)
3 kV /cm 30 kV /cm
So r 1 ā¤ 1 cm. For a given potential V , a metal object will have stronger surface fields the smaller its radius of curvature. In your microwave, the fields around the pointy ends of a fork can be large enough to ionize air, causing sparks to fly.
