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 W [See Text, Eqn. 18.1, 18.2, etc]

6. 'Hot' R = ('on' voltage)/ I = 4V/ 0.4A = 10 W

7. P = (DV) I = (4V) 0.4 A = 1.6 W

8. The KE in eV of a charge |e| accelerated through potential

difference DV is just |DV|.

9. See Section 16.4, especially Figure 16.11 (b)

10. Since I = DQ/t and the unit of charge is e, DQ=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) mF = 6/3mF . But the arrays are in parallel so

C = 6/3 + 6/3 = 4 mF

13. The total Q = CDV = 4 mF (20V) = 80 mC.

Each array stores half the total = 40 mC.

But in a series array, Q is the same for each element = 40 mC

The stored energy thus is U = Q2 /2C = 1600/12 = 133 mJ

14. DV across each array = 20V .

Connecting them in series, DV = 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 pr3

The volume of the large sphere = 8 v(small)

so 8 (4/3 pr3) = 4/3 pR3 => 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) = e0A/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 = 15mJ

17. NOTE - ERROR!! The requested ratio should be Ra/Rb!!

R = r l/A so Ra/Rb = Ab/Aa = p (22 - 1)/ p(1) = 3

18. E = k q/r2 ; V = kq/r . At r = 1m, q(enclosed) = (2 - 6) mC ,

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