Physics II

Homework X CJ

Chapter 30; 2, 8, 20, 36, 37, 42, 53, 62, 70

30.2. Visualize:

Solve: The potential difference V between two points on the y-axis is

y f V   E d y  y y i

When the electric field is uniform, the above result simplifies to V   Ey  y . In the present problem,

y  yf – yi  0.05 m – (0.05 m)  0.10 m The y-component of the electric field is

E y  5 0 , 0 0 0 V / m  V  – (–50,000 V/m)(0.10 m)  5000 V

Assess: Vf – Vi  5000 V shows that the potential at point f is higher than at point i. This is because the electric field is directed from the point at the higher potential to the point at the lower potential.

30.8. Model: The electric field is perpendicular to the equipotential lines and points “downhill.”

Visualize: Please refer to Figure Ex30.8. Three equipotential surfaces at potentials of 200 V, 0 V, and 200 V are shown.

Solve: The electric field component perpendicular to the equipotential surface is V 2 0 0 V E      2 0 , 0 0 0 V / m s 0 . 0 1 m The electric field vector is in the third quadrant, 45 below the negative x-axis. That is,  E 2 0 , 0 0 0 V / m , 4 5  b e l o w  x - a x i s 

30.20. Model: Assume that the battery is ideal and that the capacitor is a parallel-plate capacitor.

Solve: (a) From Equation 30.24, the capacitance is

1 2 2 2  A 8 . 8 5 1 0 C / N m 0 . 0 2 m  0 . 0 2 m  C 0  7 . 1  1 01 2 F  7 . 1 p F d 0 . 0 0 0 5 m (b) The charges on the electrodes are

1 2  1 0 Q  C  V C   7 . 1  1 0 F 1 0 0 V   7 . 1  1 0 C   0 . 7 1 n C 30.36. Solve: (a)

(b) Equation 30.3 gives the potential difference between two points in space: x xf x f x 2  f V  V x  V x   E d x  5 0 0 0 x V / m d x   5 0 0 0 V   2 5 0 0 x2  x 2 V  f  i x       f i  x x 2 i i   x i 2 Taking V(xi)  0 V at xi  0 m, and replacing xf with simply x, V(x)  (2500 x ) V.

(c) A graph of V versus x over the region 1 m  x  1 m is shown in part (a).

Assess: As it must be, we have d V d    2 5 0 0x2 V  5 0 0 0 x  E d x d x   x

30.37. Visualize:

Solve: Equation 30.3 gives the potential difference between two points in space

r f V  V r  V r   E d r  f  i   r r i r  r   r V r  V R   d r  l n r     l n r  l n R    l n       R   R 20r 2  0 2  0 2  R  r  V r  V 0  l n 2  0 R

Assess: At r  R, l n R R   0 and V(R)  V0, as it must be due to the assumed constant of integration.

30.42. Model: Assume the charged rod is a line of charge of length L.

Visualize: Please refer to Figure P30.42.

Solve: (a) Divide the charged rod into N small segments, each of length x and with charge q. The segment i located at position xi, contributes a small amount of potential Vi at point P: q  q  Q x L  V i    40ri 4  0 x 0 x i 4  0 x 0  x i 

Point P is at a distance x0 from the origin. This is done to avoid confusion with xi. The Vi are now summed and the sum is converted to an integral giving

L / 2 Q d x QL / 2 Q x L 2  V  l n x  x   l n 0    0   L / 2 40L L / 2 x 0 x 4  0 L 4  0 L x 0  L 2  Replacing x0 with x, the potential due to a line charge of length L at a distance x from the center is Q x L 2  V  l n   4 0 L x L 2 

(b) Because Ex   d V d x ,

Q d Q1 1  Q 1 Ex  l n x  L 2  l n x  L 2        4L d x  4  L x L 2 x  L 2 4  x2 L 2 4 0 0     0

2 Assess: When L  0 m, Ex  Q4  0 x . This is the electric field of a point charge Q a distance x away from a point charge, as expected.

30.53. Model: Assume the battery is ideal.

Visualize: The current supplied by the battery and passing through the wire is I  Vbat/R. A graph of current versus time has exactly the same shape as the graph of Vbat with an initial value of I0  (Vbat)0/R  (1.5 V)/(3.0 )  0.50 A. The horizontal axis has been changed to seconds.

Solve: Current is I  dQ/dt. Thus the total charge supplied by the battery is  Q I d t  a r e a u n d e r t h e c u r r e n t - v e r s u s - t i m e g r a p h 0 1 2 ( 7 2 0 0 s ) ( 0 . 5 0 A )  1 8 0 0 C

30.62. Visualize:

The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C1 and C2 are in parallel, their equivalent capacitance Ceq 12 is

Ceq 12  C1  C2  20 F  60 F  80 F

Then, Ceq 12 and C3 are in series. So,

1 1 1 1 1 1 9 8 0     F C e q 9  F  8 . 9  F Ce q C e q 1 2 C 3 8 0  F 1 0  F 8 0

30.70. Model: Assume the battery is ideal.

Visualize: Please refer to Figure P30.70. While the switch is in position A, the capacitors C2 and C3 are uncharged. When the switch is placed in position B, the charged capacitor C1 is connected to C2 and C3. C2 and C3 are connected in series to form an equivalent capacitor Ceq 23.

Solve: While the switch is in position A, a potential difference of V1  100 V across C1 charges it to

6 Q1 C 1 V 1 1 5  1 0 F 1 0 0 V  1 5 0 0  C When the switch is moved to position B, this initial charge Q1 is redistributed. The charge Q 1 goes on C1 and the charge Q e q 2 3 goes on Ceq 23. The voltage across C1 and Ceq 23 is the same and Q1  Q e q 2 3  Q 1  1 5 0 0  C . Combining these two conditions, we get Q  Q1 5 0 0  C  Q Q 1 e q 2 3  e q 2 3  e q 2 3 C1 C e q 2 3 C 1 C e q 2 3

1 1 1 Since C e q 2 3  2 0  F  3 0  F   1 2  F , we can rewrite this equation as 1 5 0 0  C  Q Q e q 2 3 e q 2 3  Q  667 C Q  Q  Q 1 5 0 0  C  6 6 7  C  8 3 3  C 1 5  F 1 2  F eq 23 1 1 e q 2 3

Having found the charge Qeq 23, it is easy to see that Q2  Q3  667 C because Ceq 23 is a series combination of C2 and C3. Thus,

Q 2 6 6 7  C Q 3 6 6 7  C Q 1 8 3 3  C V 2    3 3 . 3 V V 3    2 2 . 2 V V 1    5 5 . 5 V C 2 2 0  F C 3 3 0  F C 1 1 5  F