PHY 2054
Fall 2011
Exam 1 Solutions
1. Positive charges are arranged at the
vertices of an equilateral triangle as shown
(all three angles are 60o, all three sides are 1
m in length. At vertex A is 1 C of charge, at
vertex B 3 C, at vertex C 1 C. Calculate the
total of the Coulomb forces on the charge at
(A) from the charges at B and C. What is
the ratio of the magnitude of the total force
on the charge at A from both charges at B
and C in the y-direction to the magnitude of
the total force on charge A in the x direction,
|Fy| / |Fx| ?
a. 1.04
b. 0.74
c. 0.96
d. 0.87
e. 1.00
solution: 0.866 (cos30) Fab / (0.5 (sin30) Fab + Fac) = Fy/Fx where Fab = k 3*1C2/1m2 ,
Fac =k 1*1C2/1 m2, and Fac in the denominator has 100% weight since it is entirely in the xdirection. The common factors of k and C2/m2 cancel, leaving 0.866*3/(0.5*3 + 1) = 1.04
2. change vertex C to have 2 C charge
a. 0.74
b. 1.04
c. 0.96
d. 0.87
e. 1.00
solution: 0.866 (cos30) Fab / (0.5 (sin30) Fab + Fac) = Fy/Fx where Fab = k 3*1C2/1m2 ,
Fac =k 2*1C2/1 m2, and Fac in the denominator has 100% weight since it is entirely in the xdirection. The common factors of k and C2/m2 cancel, leaving 0.866*3/(0.5*3 + 2) = 0.74
3. change vertex B to have 2 C charge
a. 0.87
b. 0.74
c. 0.96
d. 1.04
e. 1.00
solution: 0.866 (cos30) Fab / (0.5 (sin30) Fab + Fac) = Fy/Fx where Fab = k 2*1C2/1m2 ,
Fac =k 1*1C2/1 m2, and Fac in the denominator has 100% weight since it is entirely in the xdirection. The common factors of k and C2/m2 cancel, leaving 0.866*2/(0.5*2 + 1) = 0.87
4. An electron , q=1.6 x 10-19 C, is accelerated in a uniform electric field of 100 N/C. What
is the acceleration of the electron, in units of 1013 m/s2?
a. 1.8
b. 0.6
c. 0.9
d. 1.2
e. 1.5
solution: F=E*q=100 N/C * 1.6 10-19 C= ma, where m of electron is 9.11 10-31 kg, so
a=100 N/C * 1.6 10-19 C/9.11 10-31 kg = 1.8 1013 m/s2
5. change field to 33 N/C
a. 0.6
b. 1.8
c. 0.9
d. 1.2
e. 1.5
c. 1.8
d. 1.2
e. 1.5
6. change field to 50 N/C
a. 0.9
b. 0.6
7. A charge of +5 C is located at the origin, and a charge of -20 C is located at x=2 m, i. e. 2
m away from the origin on the x-axis. Ignoring infinity, where along the x-axis in units of
meters is the Coulomb force equal to zero?
a. -2
b. +1
c. -1
d. -3
e. -0.67
anywhere between the charges, a positive test charge would feel a force towards the + xaxis. To the right of the -20 C at x=+2 m, a positive test charge would be pulled to the left.
Thus, only to the left of the origin, x<0, can a positive (or negative) test charge feel no force.
Ftotal = 0 = k C2 (-5/x2 + 20/(x+2)2) where the ‘-‘ indicates a force in the –x axis direction
and a ‘+’ indicates a force in the +x axis direction, and the test charge is a distance ‘x’ to
the left of the origin (the square of a minus is a plus), so that its distance from the 5 C at the
origin is x and its distance from the -20 C charge at x=+2 m is (x+2). Thus we have,
dividing the k and 5 common factors out,
0= -1/x2 + 4/(x+2)2, where the solution for ‘x’ will be positive since we have already taken
care of the fact that x is to the left of the origin. Put one term on the left to get 1/x2 =
4/(x+2)2 and take the squareroot of both sides to give 1/x=2/(x+2). The solution for x is 2 m,
where this is to the left of the origin as explained.
8. A charge of +5 C is located at the origin, and a charge of -45 C is located at x=2 m, i. e. 2
m away from the origin on the x-axis. Ignoring infinity, where along the x-axis in units of
meters is the Coulomb force equal to zero?
a. -1
b. +1
c. -2
d. -3
e. -0.67
anywhere between the charges, a positive test charge would feel a force towards the + xaxis. To the right of the -45 C at x=+2 m, a positive test charge would be pulled to the left.
Thus, only to the left of the origin, x<0, can a positive (or negative) test charge feel no force.
Ftotal = 0 = k C2 (-5/x2 + 45/(x+2)2) where the ‘-‘ indicates a force in the –x axis direction
and a ‘+’ indicates a force in the +x axis direction, and the test charge is a distance ‘x’ to
the left of the origin (the square of a minus is a plus), so that its distance from the 5 C at the
origin is x and its distance from the -45 C charge at x=+2 m is (x+2). Thus we have,
dividing the k and 5 common factors out,
0= -1/x2 + 9/(x+2)2, where the solution for ‘x’ will be positive since we have already taken
care of the fact that x is to the left of the origin. Put one term on the left to get 1/x2 =
9/(x+2)2 and take the squareroot of both sides to give 1/x=3/(x+2). The solution for x is 1 m,
where this is to the left of the origin as explained.
9. A charge of +5 C is located at the origin, and a charge of -80 C is located at x=2 m, i. e. 2
m away from the origin on the x-axis. Ignoring infinity, where along the x-axis in units of
meters is the Coulomb force equal to zero?
a. -0.67
b. +1
c. -2
d. -3
e. -1
anywhere between the charges, a positive test charge would feel a force towards the + xaxis. To the right of the -80 C at x=+2 m, a positive test charge would be pulled to the left.
Thus, only to the left of the origin, x<0, can a positive (or negative) test charge feel no force.
Ftotal = 0 = k C2 (-5/x2 + 80/(x+2)2) where the ‘-‘ indicates a force in the –x axis direction
and a ‘+’ indicates a force in the +x axis direction, and the test charge is a distance ‘x’ to
the left of the origin (the square of a minus is a plus), so that its distance from the 5 C at the
origin is x and its distance from the -80 C charge at x=+2 m is (x+2). Thus we have,
dividing the k and 5 common factors out,
0= -1/x2 + 16/(x+2)2, where the solution for ‘x’ will be positive since we have already taken
care of the fact that x is to the left of the origin. Put one term on the left to get 1/x2 =
16/(x+2)2 and take the squareroot of both sides to give 1/x=4/(x+2). The solution for x is 2/3
m, where this is to the left of the origin as explained.
10. A uniform electric field, E=100 N/C, is directed along the +x direction. If a 1 m2 area
plate is placed along the x-axis perpendicular to the electric field (see figure), what is the

total flux (E A) in units of Nm2/C ? Dashed lines are E=100 N/C uniform electric ¯eld in
the +x direction. The thick, solid black line is a 1 m2 area seen edge-on that is
perpendicular to the x- axis and to the
E ¯eld lines.
a. 100
b. 0
c. 50
d. (3/2)*100
e. 75
Soln.: 100 N/C * 1 m2 = 100 Nm2/C
11. A spherical volume of radius R contains +1 C in charge located at r=0. What is the
electric field vector magnitude at the surface of the sphere?
a. (1 C)/( 4R20)
b. (1 C)/(R 0)
c. (1 C)/(4R 0)
d. (1 C)/( R20)
e. (1 C)/ (R20)
Soln.: EA=E*4R2 ; Gauss’ Law: Qin/0 = EA so we have 1 C/0 = E4R2
So E=(4R2)-1 (1 C)/0
since surface area of sphere is 4R2
12. A cube of side L sits at the origin as shown. 1 C of charge is inside the cube at its
center. What is the total electric flux through the surface of the cube?
a. 1 C/0
b. (6 L2) * (1 C)/0
d. (6 L2) * sin45o * (1 C)/0
c. (6 L2) * cos45o * (1 C)/0
e. 0
Soln.: Gauss’ Law=> total charge inside a surface/0 = total electric flux thru the surface,
so 1C/0 is the total electric flux
13. An electric car has a battery voltage of 200 V. After 130 km of driving, the battery
needs 2*108 J of electrical energy restored to fully recharge the battery. How many
Coulombs have to be put into the positive terminal of the car battery (which remains at 200
V during the recharging process) by the battery charger? (units of Coulombs)
a. 106
b. 103
c. 104
d. 105
e. 107
Soln.: qV=2*108 J, so q=2*108/200 = 106
14. Where, in addition to infinity, in the diagram on the x-axis is the total electric force on
a positive test charge equal to zero? At x=___?___m The charge at the origin is +2 C; the
charge at x=1 m is -4 C.
x=0
x=1 m
|_____________________|
+2 C
a. -2.4
-4 C
b. -1.4
c. -0.4
d. +0.4
e. -3.4 m
Soln.: Any positive test charge between x=0 and 1m will be pushed to the right by the +2 C
and also pulled to the right by the -4 C. Any positive test charge to the right of -4 C will be
more pulled to the left by the closer -4 C charge than the further away + 2 C charge will
push to the right. Thus, the only place where the force can be zero is to the left of the +2 C
charge, call the magnitude of the distance to the left ‘x’. balance Coulomb forces:
-k(2 C)qtest/x2 (force pushes to the left) + k(4 C)qtest/(x+1m)2 (force pulls to the right) = 0
Thus, divide thru by k, qtest to get -2/x2 + 4/(x+1 m)2 =0, or 2/x2 = 4/(x+1 m)2
Take the square root of both sides: 2/x = 2/(x + 1m), so x=1/(2 – 1) = 1/0.414 = 2.4, to the
left of x=0, so answer is -2.4
14. The electric potential at the center of the square due to the + 1 nC (nanoCoulomb, or
10-9 C), + 2 nC, + 3 nC, and -1 nC charges at the corners of the square is 31.8 V. What is
the length of the edge of the square, in m?
Soln: 8.99 109 Nm2/C2 ( 1 x 10-9 nC + 2 x 10-9 nC + 3 x 10-9 nC – 1 x 10-9 nC)/(2*L/2)
where L is the length of a side, the length of a diagonal is 2 * L, and the distance from a
corner to the center is just half the length of the diagonal.
So the above expression is 8.99 * 5 * 2 /L since 10+9 in k cancels the nano in nC, so the
electric field at the center of the square is 63.6/L (units of V, where L has units of m). If
this is to be 31.8 V, then L=2 m.
15. See sketch. Points B and C each have -3q charges, while
point D has an unknown charge. Points B and C are each a
distance d from point A, while point D is a distance of 4d
away. If the electric field at point A is zero, find the charge at
point D in terms of q.
Soln: The y component of the electric field due to B and C
cancel, they are of equal magnitude and in the opposite direction. In the x direction:
𝒌
𝑬𝒙 = 𝒅𝟐 [−
𝟑𝒒
𝟏
𝑸
𝟐𝒄𝒐𝒔𝟔𝟎 + 𝟏𝟔] = 𝟎
which solves as Q = 48q.
16. Two point charges, each with charge Q, are held a distance r apart. Under these
conditions, the force between the two point charges is F. Now 1/2 of the charge Q from one
of the point charges is taken away and placed on the other point charge, without the
distance r being changed. In terms of the original force F, what is the new force between
the two point charges?
(1) (3/4)F
(2) still F
(3) (1/2)F
(4) (2/3)F
(5) (3/2)F
Soln: The new charges are Q/2 and 3Q/2. The force
between them is 3/4F.
17. The capacitances of the four capacitors shown are given
in terms of a certain quantity C. In ratio to C, what is the
equivalent capacitance between points A and D? (First
imagine that a battery is connected between those two
points; then reduce the circuit to an equivalent capacitance.).
(A) 0.85
(B) 0.41
(C) 13
(D) 0.63
(E) None of these
Soln: In the circuit with battery connected between points A and D, there are two
capacitors 2C and 6C in series (= 3/2C). Their combination is parallel to the 4C (= 11/2C)
capacitor and the last combination is in series with the C (= 11/13C = 0.85C).
18. In the above problem, now consider a battery connected between points A and B. What
fraction of the charge is stored on the 4C capacitor? Express your answer as a ratio to the
charge stored on the C capacitor.
(A) None of these
(B) 6 (C) 0.52
(D) 4.0
(E) 0.73.
Soln: In this problem, we accepted both None of these and 0.52 as the correct answers.
Now 4C and 6C are in series (=12/5C). The combination is in parallel to 2C (= 22/5C) and
the final capacitor C is in series (= 22/27C). Suppose that the battery is V volts. Then the
charge on C will be the same as the charge on the entire assembly (Q1 = 22/27CV). The
potential drop across C is 22/27V and the potential drop across the combo (4+6) is 5/27V.
It follows therefore that the charge on the combo (4+6, and therefore the capacitor 4C) is
Q4 = 12/27CV. The required ratio Q4/Q1 = 12/22 = 0.55.
19. In the circuit shown all capacitors are 6.0μF and the power supply is 5
V. The charge (in μC) on the capacitor labeled q is:
(A) 12
(B) 18 (C) 38.5
(D) 29
(E) None of these
Soln: Here we can focus on just the lower part of the circuit. Its equivalent
capacitance (from a 3 μF parallel to 6 μF = 9 μF, in series to 6 μF = 18/5 μF.
The charge in the lower part of the circuit is 18 μC. The potential drop
across the first cap is therefore V= Q/C = 3V. It follows then that the
potential drop across the capacitor q is 2V, i.e. the charge is CV = 12 μC.
20. Find the energy stored (in Joules) in the system. The values are
V = 10V , C1 = 2F, C2 = 4F, C3 = 9F, C4 = 1F, and C5 = 3F.
(1) 182
(2) 171
(3) 213
(4) 196
(5) 245
Soln: The equivalent capacitance is 40/11F and the total energy = ½ CV2 = 2000/11 = 182 J.
21-23. In the electric circuit shown, capacitors 1-4 have the same
capacitance of (2.0, 2.0, 3.0) μF each, and the amount of charge
stored in capacitor 2 is (3.0, 3.0, 4.0) μC. What is the total energy
stored in the four capacitors [in μJ]?
Answer (27, 48, 32)
Soln to 21. Capacitors 2,3 and 4 are in series and therefore must have the same charge 3μC.
Their equivalent capacitance is 2/3μF. The potential across the combo is then V = Q/C =
9/2 V. The total capacitance (with 1 in parallel to 234) = 8/3μF. Therefore the total energy
= ½ CV2 = 27 μJ.
24-26. A copper cable with resistivity ρ = 1.7 x 10-8 Ω-m is designed to carry a current of
(400, 200, 100) Amps with a power loss of 2.0 W/m. What is the required radius of this
cable (in cm)?
Answer (2.08, 1.04, 0.52)
Soln to 24: The power loss is P = I2R = I2 ρl/A. This A = πr2 = I2ρ/P/l. You have to solve
for r. When the current I = 400A, the radius is 2.08 cm. Since r is proportional to I,
reducing I would reduce r by the same factor.
27-29. A 140 pF capacitor is charged to a potential difference of 60 V, and the charging
battery is disconnected. The capacitor is then connected in parallel with a second (initially
uncharged) capacitor. If the potential difference across the first capacitor drops to (48,
52.5, 56) V, what is the capacitance of this second capacitor?
Answer (35, 20, 10) pF
Soln to 27: We start with a charge of Q = 140x60 pC on the first capacitor. In the end, we
have the same charge, a total capacitance of C+140pF and a potential difference of 48V.
Thus C = 140x60/48 – 140 = 35.
30-32. Suppose the electric company charges 10 cents per kW∙h. How much does it cost to
use a 125 W lamp (8, 6, 4) hours a day for 30 days?
Answer: ($3.00, $2.25, $1.50)
Soln for 30: In watt-hr, the total energy consumption for the month is 125x8x30 = 30000,
which is 30 kW-hr and at the rate of $ 0.1/kW-hr, costs $3.
33. Two unknown point charges Q1 and Q2 are joined by a line. At a point on the line onefourth of the way from Q1 to Q2 the electric field is zero. What is the ratio Q1/Q2?
Soln: Both charges are the same sign. You are 3 times farther from Q2, than you are from
Q1. But the field magnitude must be the same in order for the total field to cancel. Hence
Q1/Q2 = (r1/r2)2 = 1/9