Test 1, Phy2061
SPRING 2010
The electric field is given by the negative rate of change of the electric potential per
unit length (i. e. you need to take the slope at the point asked for):
E=-V/x= -400 volts/8 cm =-400 V/0.08 m= -5000
N/C at 5 cm
= 0 volts/cm (flat) at 10 cm
= -400 volts/2 cm = -20000 N/C at 14 cm
= - (-200 volts/7 cm)=+2860 N/C at 20 cm
= - (-600 volts/3 cm) = + 20000 N/C at 24 cm
F = qE = -1.602 10-19 C * E values above. Since the charge is an electron,
the sign should change.
2. Three equal
charges +q are
placed at the
vertices of an
equilateral
triangle, side of
length ‘L’.
Three points, labeled a, b, c are shown in the figure. As shown: ‘a’ is midway
between the leftmost and top charges, ‘b’ is equidistant from all three
charges(lies on the intersection of the bisectors of all three 60o angles), ‘c’ is just
as far below the line connecting the bottom two charges as ‘b’ is above this line,
and at the same x-coordinate. Calculate the electric field at each of the three
points (magnitude and both the x- and y-components ); indicate with a vector at
each of the three points what this E field looks like. (direction and magnitude
both count.) At which of the three points is the E field the largest in magnitude?
At ‘a’, the left charge and the top charge sum to Ex = Ey = 0, just leaving the field
from the right hand charge. The magnitude of the E field from this charge at ‘a’
is (q/40)(1/0.866L)2 = 1.333(q/40L2). The x-component is then (- cos30) times
this (Ex= -(q/40L2)(1.155) and the y-component is + sin30 times the magnitude,
or Ey = + 0.667(q/40L2)
At points b and c, there is no x-component of E from symmetry. ‘b’ is 0.289 L
above the base (this distance = tan30 * L/2) and 0.578 L from each of the +
charges (this distance is L/2 * cos30). Thus, the Ey at b from the two charges on
the base (the sin30 gives the y-component, the factor of 2 is from there being two
charges) is 2*sin30(q/40)(1/0.577 L)2 = +(q/40)(3/L2). The Ey component at b
from the upper charge is –(q/40)(1/0.577L)2 (where the 0.577 is the distance of
‘b’ from each vertex of the triangle), so these contributions sum to 0 at b. At ‘c’,
the contribution from the two charges on the base is Ey= - 3q/(40L2) and from
the top charge Ey= - (q/40)/(1.155L)2 (the height of an equilateral triangle of
side L is (3)1/2/2 L = 0.866 L so the distance from the top charge to c is 0.866 +
0.289 L) so the sum at ‘c’ is Ey= - (q/40L2)(3 + 0.75) = - (3.75q/40L2).
So the largest field in magnitude is at ‘c’ where symmetry is not cancelling some
(or all) of the fields from the three charges.
3. A continuous, homogeneous distribution of positive charge in the x-y plane (z
axis is perpendicular out of the page) has the shape shown (sort of like a
racetrack in shape), with two straight parallel sides each of length ‘L’ which are
a distance of 2 R apart and two semi-circle shaped ends each of radius ‘3 R’
centered at the origin in the middle, with L=4√2 R. If the total charge on the
‘racetrack’ shaped insulator is Q
a.) what is the lineal charge density  in terms of Q and R?
total length = 2 * 4√2 R + 2 * arclength at end, where the angle subtended by
each arc is  and sin/2 = R/3R. so =38.94 o so two times this (length total of
both end arcs) is 77.88o, which is 77.88/360 * circumference of a circle with
radius 3 R, so the total arc length is 4.10 R , so =Q/(15.4 R)
b.) what is the electric field at a distance z=6R above the middle of the track
(the middle is at the origin of the x-y axes)? Give all the components (x, y,
and z)
the contributions from the two straight sides (one of which is the black dot in
the drawing just above) cancel in the x and y directions, and (each) has the zcomponent result from a straight line of charge
eq. 26-16 E=(1/40)  L/(y(y2 + L2/4)1/2) where L=4(2)1/2 R here and
at a distance y of (37)1/2 * R, all times sin Θ=6/(37)1/2 = 0.986 (this gives the
component of the E field between the straight side and the point 6 R above
the center of the shape in the z direction), where tanΘ=6 (the base of the
triangle is from the origin to one straight side, distance R, and the height of
the triangle if 6 R). So
Efrom 2 straight sides =
2 * 0.986 * (1/40)  (4(2)1/2 R)/((37)1/2R*(37 R2 + 8R2)1/2)=
2*0.986*(1/40)(Q/R2) 5.66/(15.4*6.083 *6.71) where 15.4 comes from ,
(37)1/2=6.083 and 6.71 = (37 + 8)1/2, so
Efrom 2 straight sides=(1/40)*0.00900 Q/R2
Efrom the semi-circle ends also is only in the z-direction, and is the result from a ring,
but not the full 2 radians but just (see part a.) 77.88/360 * 2.
The result for z above a full ring of radius R is
E=z(2R)/(40 (z2 + R2)3/2) so we use  from part a., z=6R, 2R ->
(77.88/360)*2(3R) so Etotal=
(1/40) * ( 0.01776 Q/R2 + Q*6R*4.078R/(15.4R*(36R2+9R2)3/2))
=(1/40) * ( 0.01776 Q/R2 + (Q/R2) *0.00525)=0.02301 Q/R2 *(1/40)
4. Two square parallel plates has a uniformly distributed charge density of
3 x 10-4 C/m2 with one plate positively charged and one negatively charged. The
space between the plates is filled with a material with dielectric constant 10.
Considering locations near the center of the plates (where the fields are uniform),
find
a.) The net electric field in the dielectric material
E0=/0 = 3 10-4 C/m2/8.85 10-12 C2/Nm2 = 0.339 108 N/C = 3.39 107 N/C
So net field, Enet, is E0/e=3.39 106 N/C
b.) The induced charge density (units of C/m2) on the surfaces of the dielectric
material.
Polarization field due to the dielectric is E’ = E0 – Enet =( 33.9 – 3.39 ) 106N/C = 30.51
106 N/C, so that the induced charge that created E’ is ind=0E’=8.85 10-12 C2Nm2 *
30.51 106 N/C = 270.0 10-6 C/m2 = 2.70 10-4 C/m2
5. Which arrangement in the figure has the largest capacitance? Show the value for
C for each of the three. All plates have a uniform distribution of charge with total
charges of +q and –q; you may ignore fringe fields on the edges of the plates. In a.),
two circular plates, both with radius=R, are separated by 0.1 R In b.), two square
plates of side 2R have separation 0.125 R. In c.), two equilateral plates with sides of
length 3R have separation 0.1 R
for a., C=0A/d where A=R2 and d=0.1 R, so C=0 R * 10 = 31.4 R 0
for b., C=0 4R2/(1/8 R) = 32 R 0
for c., C=0 1/2base*height/0.1R = 0 (1.5 R) (√3/2 3R)/0.1 R = 39.0 R 0
6. A half-ring of uniformly distributed total charge –Q and radius R is centered at
the point marked by an ‘o’ in the figure, which is at the origin of the x,y,z coordinate
frame.
a. What is the lineal charge density ?
= - Q/R
In the plane of the half ring, what is
b. Ex
dEx= dq cosΘ/40R2 where Θ varies between 90o (/2) at the top of the halfring, to 0o at the left side, to - /2 at the bottom of the ring. The cosΘ gives
the x-component, and the sign would be in the +x direction if dq were
positive, and is therefore in the negative direction as it should be.
dq=(-Q/R)ds where the element of arc length ds=RdΘ
Ex = integral from -/2 to +/2 (-Q/R)R (cosΘ dΘ/40R2) =
-Q/420R2 (sinΘ) evaluated between -/2 to +/2 = -Q/220R2
c. Ey =0 by symmetry
c. Ez =0 by symmetry
7. The 30 V battery has fully charged the 7 µF capacitor in the circuit shown. The
battery is now removed and points A and B are connected together by a shorting
wire.
a.) draw a graph of the Voltage across the capacitor, VC, as a function of time.
What voltage does it start from at t=0 just as the discharge starts?
b.) At what time will the voltage across the capacitor be 10 V? (3 sig figs please.)
Solution: a.) for the graph, see Fig. 31-22 a. VC(t=0) = 30 V (fully charged)
b.) eq. 31-32 tells us q = q0 e**(-t/RC), and q=CV, so V(t)=q(t)/C So if V(t?) is to be
1/3 of its value at t=0, we need e**(-t?/RC) = 1/3. So ln(1/3) = -1.0986 = (-t?/RC)
and t? = 1.0986 *1000 * 7 10-6 = 7.69 ms.
8. You have seven 10  resistors arranged as shown in the figure.
a.) What is the net resistance between points A and B with this arrangement of
resistors?
Solution: the two parallel arrangements have 5  and 3.33  respectively (since
1/Rparallel = 1/R1 + 1/R2 + … 1/Rn and Rparallel for the first arrangement is (1/10 +
1/10)-1 = (0.2)-1 = 5 and for the second arrangement (1/10 + 1/10 + 1/10)-1 = (0.3)-1 =
3.33)
Thus, RAB = 10 + 5 + 3.33 + 10  = 28.33 
b.) What arrangement of all 7 resistors between A and B gives RAB=46.666 ?
Draw a legible picture and show your calculation.
Solution: arrange 4 resistors in series, and the other 3 as shown