Solutions to PHY2054 Exam 1, 11 June 2002
1. Draw the force vectors at x. F(-q) is attractive to left & upward;
F(+q), repulsive to right & upward. The resultant is up = +y.
2. Draw the force vectors on q as in #1, above. The resultant is to the
right and parallel to a.
3. Sketch in the E-field lines for the (Q, -Q) system. A line drawn
from q to the center of a is perpendicular to the field lines, so it must
be an equipotential.
Thus no work is done in moving along this path so W = 0!
4. Draw the force vectors or you won't follow the argument!
If q = 2Q, the force due to +Q must be F(+Q) = k [Q(2Q)]/(a cos 60)2 .
But cos 60 = ½, so (a cos 60)2 = a2/2 ;
thus F(+Q) = 4kQ2/a2 , directed upward at 60 to the horizontal.
The force due to (-Q) has same magnitude, directed downward at 60 .
Adding these vectors,
The resultant force thus is 2 [4kQ2/a2] cos 60 = 4kQ2/a2
5. The internal r is just the (on-off potential difference) / (current)
= (5.6V - 4V)/ (0.4 A) = 4 
[See Text, Eqn. 18.1, 18.2, etc]
6. 'Hot' R = ('on' voltage)/ I = 4V/ 0.4A = 10 
7. P = (V) I = (4V) 0.4 A = 1.6 W
8. The KE in eV of a charge |e| accelerated through potential
difference V is just |V|.
9. See Section 16.4, especially Figure 16.11 (b)
10. Since I = Q/t and the unit of charge is e, Q=ne,
so n = It/e = 2 x 2C/s/ e = 2.5 x1019
11. Sketch the force vectors: p2 repels p1 (=left); e attracts p1 (=right),
but e is farther away so the resultant is (left).
12. Top & bottom arrays are each in series so the C of each array is
1/(1/6 +1/6 + 1/6) F = 6/3F . But the arrays are in parallel so
C = 6/3 + 6/3 = 4 F
13. The total Q = CV = 4 F (20V) = 80 C.
Each array stores half the total = 40 C.
But in a series array, Q is the same for each element = 40 C
The stored energy thus is U = Q2 /2C = 1600/12 = 133 J
14. V across each array = 20V .
Connecting them in series, V = 20V + 20V = 40V.
No circuit was completed so the charges remain in place and
the stored energy is unaffected.
15. We have small spheres coalescing to form a large sphere.
Sphere volume = 4/3 r3
The volume of the large sphere = 8 v(small)
so 8 (4/3 r3) = 4/3 R3 => R = (8)1/3 r = 2 r
The potential at a sphere's surface is just kq/r, so for the large
sphere, V = k (8q)/2r = 4 kq/r = 4V
16. C (parallel-plates) = 0A/d . If d is doubled, C is halved.
Stored energy U = Q2/2C and Q is unchanged.
The work done in doubling d is the difference = 100/4 -100/10 = 15J
17. NOTE - ERROR!! The requested ratio should be Ra/Rb!!
R =  l/A so Ra/Rb = Ab/Aa =  (22 - 1)/ (1) = 3
18. E = k q/r2 ; V = kq/r . At r = 1m, q(enclosed) = (2 - 6) C ,
so E= 9e9 (4e-6) = 36 e3 V/m. Similarly, V = 36 KV
19. At r = 6 cm, we are inside the conducting shell; thus E = 0
20. Ar r =6 cm, the potential is same as at outer surface
= k Q(enc)/r = 9e9 (4e-6)/ (7e-2) = 5e5 V