1. Suppose that two point charges, each with a charge of +1.00 Coulomb are separated by a distance of 1.00 meter. Determine the magnitude of the electrical force of repulsion between them. Given:
Find:
Q1 = 1.00 C
Felectronic force = ?
Q2 = 1.00 C d = 1.00 m Solution : Felectronic force = k • Q1 • Q1 • Q2 / d2 Felectronic force force = (9.0 x 109 N•m2/C2) 109 N•m2/C2) • (1.00 C) • (1.00 C) / (1.00 m)2 Answer : Felectronic force = 9.0 x 109 N
2. Two balloons are charged with an identical quantity and type of charge: -6.25 nC. They are held apart at a separation distance of o f 61.7 cm. Determine the magnitude of the electrical force of repulsion between them. Given:
Find:
Q1 = -6.25 nC = -6.25 x 10-9 C
Felectronic force = ?
Q2 = -6.25 nC = -6.25 x 10-9 C d = 61.7 cm = 0.617 m Solution : Felect = k • Q1 • Q1 • Q2 / d2 Felect = (9.0 x 109 N•m2/C2) 109 N•m2/C2) • (6.25 x 10-9 10-9 C) • (6.25 x 10-9 10-9 C) / (0.617 m)2 Answer : Felectronic force = 9.23 x 10-7 N
3. Two balloons with charges of +3.37 µC and -8.21 µC attract each other with a force of 0.0626 Newton. Determine the separation distance between the two balloons. Given:
Find:
Q1 = +3.37 µC = +3.37 x 10-6 C
d=?
Q2 = -8.21 µC = -8.21 x 10-6 C Felect = -0.0626 N Solution : Felectronic force = k • Q1 • Q1 • Q2 / d2 d2 • Felectronic Felectronic force = k • Q1 • Q1 • Q2 d2 = k • Q1 • Q1 • Q2 / Felect d = SQRT(k • Q1 • Q1 • Q2) / Felectronic force d = SQRT [(9.0 x 109 N•m2/C2) 109 N•m2/C2) • (-8.21 (-8.21 x 10-6 C) • (+3.37 x 10-6 10-6 C) / (-0.0626 N)] d = Sqrt [ +3.98 m2 ] Answer : d = +1.99 m
4. Find the repulsive Coulomb force between two protons in a nucleus of iron?
Distance is 4x10-15 meter. Given: We know that chrage of roton q1 =q2= 1.6 x 10-19 C Here distance, r = 4 x 10 -15
Solution:
F = (1/4
0)
/ (q1q2/r 2)
F = 9 x 109 x (1.6 x 10-19)2 /(4 x 10-15)2 Answer: F = 14 N.
5. The distance between the proton and the electron in the hydrogen is r = 5.3 x 10 -11 meter. What is the magnitude force for two particles? Given: q1 = q2 = 1.6 x 10-19C r = 5.3 x 10-11 meter Solution:
F = (1/4
0)
/ (q1q2/r 2)
F = 9 x 109 x (1.6 x 10-19)2 /(5.3 x 10-11)2 Answer: F = 8.1 x 10-8 N
6. A particle charged with 20C placed 50cm from another 70C charged particle. Find the force between them. Given, q1 = 20C q 2 = 70C Distance, r = 50cm = 0.5m Force, F =? Solution: F = (1/4
0)
/ (q1q2/r 2)
F = 9 x 109 x 20 x 70 /(0.5)2 Answer: F = 5.04 x 1013 N
7. A point charge q1 has a magnitude of 3 x 10-16C. A second charge q2 has a magnitude of -1.5 x 10-6C and is located 0.12m from the first charge. Determine the electrostatic force each charge exerts on the other. Solution: F =kqq/r 2 F = (9 x 109 )x (3 x 10-6) (1.5 x 10-6) /( 0.12)2 Answer: F = 2.81 N
8. The force between two identical charges separated by 1 cm is equal to 90 N. What is the magnitude of the two charges? Given: F = 90 N q1 = charge of first body q2 = charge of second body r = 1 cm Solution : Use the Coulomb’s Law equation F = (k) ((q1q2/r 2)) The problem says the two charges are identical, so q1 = q2 = q Substitute this into the equation F = (1/4
q=
0) / (q1q2/r2)
m) √ .(999 xN)(.9 Nm/C
Answer: q = ±1.00×10-6 Coulombs
9. Two point charges, QA = +8 C and QB = -5 C, are seperated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 109 Nm2C-2. Given:
QA = +8 C QB = -5 C k = 9 x 109 Nm2C-2 Solution: F = (k) ((q1q2/r 2)) F = (8 x 10-6 )x (5 x 10-6) /( 0.1)2 F = (40 x 10-12)/(0.1)2 = 40 x 10-12 /(0.01)= 4000 x 10-12 Answer: F = 4 x 10-9 N
10. Two charged particles as shown in figure below, Q p = +10 C and QB = +20 C, are seperated by a distance r = 10 cm. What is the magnitude of the electrostatic force. Given:
Q p = +10 C QB = +20 C k = 9 x 109 Nm2C-2 Find: Electric force Solution: F = (k) ((q1q2/r 2)) F = (9 x 109)(((10 x 10-6)(20 x 10-6))/(12 x 10-2)2) = (9 x 109) (200 x 10-12)/(144 x 10-4) Answer: F = 125 N
11. What is the strength and direction of the electric field 3.74 cm on the left hand side of a 9.1 mC negative charge?
Given:
E=? K = 9.0 x 10 9 N . m 2 / C 2 d = 3.74 cm q = 9.1 mC
Find:
E is the electric field strength in N / C
Solution:
E = ( 9.0 x 10 9 N . m 2 / C 2 ) ( 9.1 x 10 -6 C ) / ( 3.74 x 10 -2 m ) 2 E = 5.8552 x 107 N / C
Answer:
E = 5.9 x 107 N / C ; to the right
12. At what distance from a negative charge of 5.536 nC would the electric field strength be 1.90 x 105 N/C ?
Given:
d=? q = 5.536 nC E = 1.90 x 105 N / C K = 9.0 x 10 9 N . m 2 / C 2
Find:
d is the distance in meters ( m ) away from the charge.
Solution:
d=
√
d=
(9. x )(5.53 x ) .9 x
d = square root of 2.6223 x 10 -4 m 2 d = 1.6194 x 10 -2 m Answer: d = 1.6 cm
13. If it takes 88.3 J of work to move 0.721 C of charge from a positive plate to a negative plate, what is the potential difference (voltage) between the plates?
Given:
W = 88.3 J q = 0.721 C V=?
Solution: V=
V = 88.3 J 0.721 C
V = 122.46879 v
Answer:
V = 122 v
14. Two parallel oppositely charged plates are 5.1 cm apart. The potential difference, in volts, between the plates is 44.6 v. Find the electric field strength between them.
Given:
d = 5.1 cm V = 44.6 v E=?
Solution:
V=Ed
E = V/d
E = 44.6 v/ 5.1 x 10 -2 m
E = 8.745098 x 10 2 N / C
Answer:
E = 8.7 x 10 2 N / C
15. A charge is in the electric field. If a force of 20 N is applied to it has charge value as 10 μμ C, calculate the electric field intensity. Given: Force F = 20 N, charge q = 10 μμ C Solution : E = FqFq E = 20N/10×10−6C Answer: E = 2 × 106 N/C
16. What is the electric field intensity if a force of 12 N is applied to the charge which is in the electric field having the charge value as 36 μμ C. Given: Force F = 12 N, charge q = 36 μ C Solution: E = F/q E = 12N/36×10−6C Answer: E = 36 × 104 N/C.
17. A uniform line charge of infinite extent with (6,8,3)m. Solution:
√ + = √ 6 +8^2 =10
E=
r c =
∴
E =
^−9 (.5)
Answer: E = 36 r c V/m
=
/
lies on z-axis. Find
at
18. What is the electric field strength at a distance of 1 0 cm from a charge of 2 μC?
Solution:
E=
=
= (9 ()() ) =1.8 105
N/C
Answer:
= 1.8 105
N/C
19. What is the magnitude of a point charge whose E-field at a distance of 25 cm is 3.4 N/C Solutions: E = kq/r 2 3.4 = (9 x 109)q/(0.25)2 Answer: Q = 2.4 x 10-11C
20. A small charge Q = 6.0 mC is found in a uniform E-field. Determine the force on the charge.
Given: Q = 6.0 mC E = 2.9 N/C Find: F=? Solution: F = qE Answer: F = (6 x 10-3)(2.9) = 0.02 N
21. A closed surface encloses a net charge of 10 nC . What is the net electric flux through the surface? Solution:
Answer:
22. Three small spheres are enclosed in surface carrying charges q1 = 4 nC, q2 = -10.00 nC, and q3 = 5 nC. Surface (s)
What it encloses
s1
q1
s2
q2
s3
q1 and q2
s4
q1 and q3
s5
q1, q2 and q3
Find the net electric flux through each of the following closed surfaces. (a) s1 (b) s2 (c) s3 (d) s4 (e) s5 Solution and Answer:
(a)
(b)
(c)
(d)
(e)
23. An electric field of 500 V/m makes an angle of 30.00 with the surface vector, which has a magnitude of 0.500 m2. Find the electric flux that passes through the surface. Solution: ΦE = E • A = EA cosθ ΦE = (500 V/m)(0.500 m2)cos30.00 Answer: ΦE = 217 V m
24. Electric field of a spherical charg e distribution. A spherical charge distribution of 7.52 × 10−5 C has a radius of 5.25 × 10−3 m. Find the electric field intensity E at (a) 5.00 cm from the charge and Solution: a. E = k (q/ r2) E = (9.00 x 109 Nm2/C2)(7.52 x 10−5C)/ (0.050 m)2 Answer: E = 2.71 × 108 N/C
25. Electric field of a spherical charg e distribution. A spherical charge distribution of 7.52 × 10−5 C has a radius of 5.25 × 10−3 m. Find the electric field intensity E at 3.25 × 10 −3 m inside the charge. Solution: E = k (r/R 3) q E = (9.00 x 102 Nm2/C2)(( (3.25 x 10−3C)/(5.25 x 10-3 m)3)(7.5 x 10-5 C) Answer: E = 1.52 × 1010 N/C
26. Electric field of an infinite line of charge. An infinite line of charge carries a linear charge density of 2.55 × 10−5 C/m. Find the electric field E at a distance of 15.5 cm from the line of charge.
Solution:
/
E = 2k r E = (2(9.00 x 109 Nm 2/C2)(2.55 x 10−5C/m) )/0.155 m Answer: E = 2.96 × 106 N/C
27. Electric field of an infinite sheet of charg e. Find the electric field in front of an infinite sheet of charge carrying a surface charge density of 3.58 × 10−12 C/m2. Solution : E=
/2
o
E = (3.58 x 10−12C/m2 )/2(8.85 x 10−12C2/Nm 2) Answer: E = 0.202 N/C
28. The electric field between two opositely charged conducting plates. Find the electric field between two opositely charged circular conducting plates o f 5.00 cm radius, if a charge of 8.70 × 10−10 C is placed on the plates. Solution: A = πr 2 = π(0.0500 m)2 = 7.85 × 10−3 m2 E=
/
oA
E = (8.70 x 10−10C) /(8.85 x 10−12C2/Nm 2)(7.85 x 10−3m2) Answer: E = 1.25 × 104 N/C
29. A cubical volume of edge a in region where an electric field exist which depends on x coordinate as E = (ax2 + b/2) V/m Find the amount of charge enclosed in the cubical volume. Solution
E2 a2 – E1a2 =
q=
o
a2 (E2 – E1) =
Answer: = 3 o a5
o
o
a2 (4a3 -a3)
30. Three small spheres are enclosed in surface carrying charges q1 = 4 nC, q2 = -10.00 nC, and q3 = 5 nC. Surface (s)
What it encloses
s1
q1
s2
q2
s3
q1 and q2
s4
q1 and q3
s5
q1, q2 and q3
Find the net electric flux through each of the following closed surfaces. s1 s2 s3 s4 s5 Solution and Answer:
31. There are three charges q1, q2, and q3 having charge 6 C, 5 C and 3 C enclosed in a surface. Find the total flux enclosed by the surface.
Solution: Total charge Q, Q = q1 + q2 + q3 =6C+5C+3C = 14 C The total flux, ϕ = Q/ϵ0 ϕϕ = 14C/8.854×10-12 m3 / kg/s4 A2
Answer: ϕϕ = 1.584 Nm2 / C−1
32. Find the electric flux in a cylinder having length 5 cm, radius 2 cm having electric field intensity 2 N/C. Given: length l = 5 cm radius r = 2 cm, Electric field intensity E = 2 N/C Solution: Electric flux, ϕ= E (2 π rl) ϕ = 2 N/C (2 π x 0.02 m ×0.05 m) Answer: ϕ = 0.0125 Nm2 / C−1
33. A 4.0 cm-square in the x-y plane sits in a uniform electric field E = (2.0 I +3 .0 j + 5.0k) N/C. Find the electric flux through the square.
Solution:
ϕE = E perpendicular to A = ( 5.0 N/C) x (16 cm2)
Answer: ϕE = 8.0 x 10-3 Nm2/C
34. A thundercloud produces a vertical electric field of magnitude 28.0 kN/C at ground level. You hold a 22.0 cm x 28.0 cm sheet of paper horizontally below the cloud. (a) What is the electric flux through the sheet? Solution: ϕE = E A = ( 28.0 x 103 N/C) x (0.0616 m2)
Answer:
ϕE = 1.72 x 103 Nm2/C
35. An infinite uniform line charge ρL = 2 nC/m lies along the x axis in free space, while point charges of 8 nC each are located at (0,0,1) and (0,0,-1).
a) Find E at (2,3,-4).
Solution:
Answer:
36. The portion of the z axis for which |z| < 2 carries a nonuniform line charge density of 10|z| nC/m, and ρL = 0 elsewhere. Determine E in free space at:
a) (0,0,4): The general form for the differential field at (0,0,4) is
Solution:
Answer:
37. Calculate the divergence of D at the point specified if D = (1/z2) 10xyz ax + 5x2z ay + (2z3 − 5x2y) az at P(−2, 3, 5): Solution:
∇ ∇
· D = (10y/ z +0+2+ 10x2y/ z3)
Answer: · D = = 8.96
(−2,3,5) =
8.96
38. An electric field is given as E = −10ey(sin 2z ax + x sin 2z ay + 2x cos 2z az) V/m.
Find E at P(5, 0, π/12):
Solution:
EP = −10 [sin(π/6) ax + 5 sin(π/6) ay + 10 cos(π/6) az] Answer:
39. A point charge of 20 nC is located at (4,-1,3), and a uniform line charge of -25 nC/m is lies along the intersection of the planes x = −4 and z = 6. Find: Calculate D at (3,-1,0) Solution:
Answer:
40. Volume charge density is located in free space as ρv = 2e−1000r nC/m3 for 0 < r < 1 mm, and ρv = 0 elsewhere. Find the total charge enclosed by the spherical surface r = 1 mm Solution:
Answer:
41. A uniform volume charge density of 80 µC/m3 is present throughout the region 8 mm
Answer:
42. Three vectors extending from the origin are given as r1 = (7, 3, −2), r2 = (−2, 7, −3), and r3 = (0, 2, 3). Find: Solution: a unit vector perpendicular to both r1 and r2: Solution:
a unit vector perpendicular to the vectors r1 −r2 and r2 −r3: r1 −r2 = (9, −4, 1) and r2 −r3 = (−2, 5, −6). So r1 − r2 × r2 − r3 = (19, 52, 32).
43. Two point charges of Q1 coulombs each are located at (0,0,1) and (0,0,-1). (a) Determine the locus of the possible positions of a third charge Q2 where Q2 may be any positive or negative value, such that the total field E = 0 at (0,1,0):
Solution:
44. Point charges of 50nC each are located at A(1, 0, 0), B(−1, 0, 0), C(0, 1, 0), and D(0, −1, 0) in free space. Find the total force on the charge at A.
Solution:
Answer:
45. A 100 nC point charge is located at A(−1, 1, 3) in free space. Find the locus of all points P (x, y, z) at which Ex = 500 V/m: The total field at P will be: Solution:
Answer:
46. Charges of 20 and -20 nC are located at (3, 0, 0) and (−3, 0, 0), respectively. Let = 0. Determine |E| at P (0, y, 0): The field will be
Solution
Answer:
47.The volume charge density ρv = ρ0e−|x|−|y|−|z| exists over all free space. Calculate the total charge present Solution:
Answer:
48. A uniform volume charge density of 0.2 µC/m3 (note typo in book) is present throughout the spherical shell extending from r = 3 cm to r = 5 cm. find the total charge present throughout the shell: Solution:
Answer: 82.1 pC
49. Uniform line charges of 0.4 µC/m and −0.4 µC/m are located in the x = 0 plane at y = −0.6 and y = 0.6 m respectively. Let = 0. Find E at P (x, 0, z): Solution:
Answer:
