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19/10/2020G

Electric Fields

Ch. 23

Phys 104 - Ch. 23/IIII - lecture 4

1442 - 1st semester

Dr. Ayman Alismail

23Ch.

𝐸1 = 𝑘𝑒𝑞1

𝑟12

𝐸1 = 8.99 × 109 ×

−3.00 × 10−9

0.100 2

𝐸1 = 2.7 × 103 ΤN C

𝐸2 = 𝑘𝑒𝑞2

𝑟22

𝐸2 = 8.99 × 109 ×

6.00 × 10−9

0.300 2

𝐸2 = 5.99 × 102 ΤN C

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 2

23.04sec.

Problem 23.19

Three point charges are arranged as shown in the following figure. (a) Find the vector electric field that the 6.00 nC and −3.00 nC charges together create atthe origin. (b) Find the vector force on the 5.00 nC charge.

The Electric Field

(a)

𝐸𝑥 = 𝐸1𝑥 + 𝐸2𝑥

𝐸𝑥 = 0 + 𝐸2 cos(𝜃2)

𝐸𝑥 = 5.99 × 102 × cos(180)

𝐸𝑥 = −5.99 × 102 ΤN C

𝐸𝑦 = 𝐸1𝑦 + 𝐸2𝑦

𝐸𝑦 = 𝐸1 sin(𝜃1) + 0

𝐸𝑦 = 2.7 × 103 × sin(270)

𝐸𝑦 = −2.7 × 103 ΤN C

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 3

23.04sec.

Problem 23.19

Three point charges are arranged as shown in the following figure. (a) Find the vector electric field that the 6.00 nC and −3.00 nC charges together create atthe origin. (b) Find the vector force on the 5.00 nC charge.

The Electric Field

(a)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 4

23.04sec.

Problem 23.19

Three point charges are arranged as shown in the following figure. (a) Find the vector electric field that the 6.00 nC and −3.00 nC charges together create atthe origin. (b) Find the vector force on the 5.00 nC charge.

The Electric Field

𝐸 = 𝐸𝑥2 + 𝐸𝑦

2

𝐸 = −5.99 × 102 2 + −2.7 × 103 2

𝐸 = 2.77 × 103 ΤN C

𝜑 = tan−1𝐸𝑦

𝐸𝑥

𝜑 = tan−1−2.7 × 103

−5.99 × 102

𝜑 = 257.49° !!

or

𝐸 = −5.99 × 103 Ƹ𝑖 − 2.7 × 103 Ƹ𝑗 ΤN C

(a)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 5

23.04sec.

Problem 23.19

Three point charges are arranged as shown in the following figure. (a) Find the vector electric field that the 6.00 nC and −3.00 nC charges together create atthe origin. (b) Find the vector force on the 5.00 nC charge.

The Electric Field

(b) 𝐹3 = 𝑞3𝐸

𝐹3 = 5.00 × 10−9 × 6.57 × 103

𝐹3 = 3.29 × 10−5 N

𝜑 = tan−1𝐹3𝑦

𝐹3𝑥

𝜑 = 204.26°

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 6

23.04The Electric Fieldsec.

Problem 23.21

Four point charges are at the corners of a square of side a as shown in the following figure. (a) Determine the magnitude and direction of the electric field at

the location of charge 𝑞. (b) What is the resultant force on 𝑞?

𝐸1 = 𝑘𝑒𝑞1

𝑟12

𝐸1 = 𝑘𝑒2𝑞

𝑎2

𝐸2 = 𝑘𝑒𝑞2

𝑟22

𝐸2 = 𝑘𝑒3𝑞

2𝑎2

𝐸2 = 𝑘𝑒3𝑞

2𝑎2

(a)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 7

23.04The Electric Fieldsec.

Problem 23.21

Four point charges are at the corners of a square of side a as shown in the following figure. (a) Determine the magnitude and direction of the electric field at

the location of charge 𝑞. (b) What is the resultant force on 𝑞?

𝐸3 = 𝑘𝑒𝑞2

𝑟32

𝐸3 = 𝑘𝑒4𝑞

𝑎2

(a)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 8

23.04The Electric Fieldsec.

Problem 23.21

Four point charges are at the corners of a square of side a as shown in the following figure. (a) Determine the magnitude and direction of the electric field at

the location of charge 𝑞. (b) What is the resultant force on 𝑞?

𝐸𝑥 = 𝐸1𝑥 + 𝐸2𝑥 +𝐸3𝑥

𝐸𝑥 = 𝑘𝑒2𝑞

𝑎2+ 𝑘𝑒

3𝑞

2𝑎2cos 45 + 0

𝐸𝑥 = 3.06 ×𝑘𝑒𝑞

𝑎2

𝐸𝑦 = 𝐸1𝑦 + 𝐸2𝑦 +𝐸3𝑦

𝐸𝑦 = 0 + 𝑘𝑒3𝑞

2𝑎2sin(45) + 𝑘𝑒

4𝑞

𝑎2

𝐸𝑦 = 5.06 ×𝑘𝑒𝑞

𝑎2

(a)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 9

23.04The Electric Fieldsec.

Problem 23.21

Four point charges are at the corners of a square of side a as shown in the following figure. (a) Determine the magnitude and direction of the electric field at

the location of charge 𝑞. (b) What is the resultant force on 𝑞?

𝐸 = 𝐸𝑥2 + 𝐸𝑦

2

𝐸 = 3.06 ×𝑘𝑒𝑞

𝑎2

2

+ 5.06 ×𝑘𝑒𝑞

𝑎2

2

𝐸 = 5.91 ×𝑘𝑒𝑞

𝑎2

𝜑 = tan−1𝐸𝑦

𝐸𝑥

𝜑 = tan−15.06 ×

𝑘𝑒𝑞𝑎2

3.06 ×𝑘𝑒𝑞𝑎2

𝜑 = 58.83°or

𝐸 = 3.06 Ƹ𝑖 + 5.06 Ƹ𝑗𝑘𝑒𝑞

𝑎2

(a)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 10

23.04The Electric Fieldsec.

Problem 23.21

Four point charges are at the corners of a square of side a as shown in the following figure. (a) Determine the magnitude and direction of the electric field at

the location of charge 𝑞. (b) What is the resultant force on 𝑞?

𝐹4 = 𝑞4𝐸

𝐹4 = 𝑞 × 5.91 ×𝑘𝑒𝑞

𝑎2

𝐹4 = 5.91 ×𝑘𝑒𝑞

2

𝑎2

𝜑 = tan−1𝐹4𝑦

𝐹4𝑥

𝜑 = 58.83°

(b)

The electric field at any point 𝑥 is:

𝐸𝑥 = 𝐸1𝑥 + 𝐸2𝑥

𝐸𝑥 = −𝑘𝑒−𝑞

𝑥 + 𝑎 2+ 𝑘𝑒

𝑞

𝑥 − 𝑎 2

𝐸𝑥 = 𝑘𝑒𝑞

𝑥 − 𝑎 2− 𝑘𝑒

𝑞

𝑥 + 𝑎 2

𝐸𝑥 = 𝑘𝑒𝑞𝑥 + 𝑎 2 − 𝑥 − 𝑎 2

𝑥 − 𝑎 2 𝑥 + 𝑎 2

𝐸𝑥 = 𝑘𝑒𝑞𝑥2 + 2𝑎𝑥 + 𝑎2 − 𝑥2 − 2𝑎𝑥 + 𝑎2

𝑥 − 𝑎 𝑥 + 𝑎2

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 11

23.04sec.

Problem 23.22

Consider the electric dipole shown in the following figure. Show that the electric field at a distant point on the +𝑥 axis is 𝐸𝑥 ≈ 4𝑘𝑒𝑞𝑎

𝑥3.

The Electric Field

𝐸𝑥 = 𝑘𝑒𝑞𝑥2 + 2𝑎𝑥 + 𝑎2 − 𝑥2 + 2𝑎𝑥 − 𝑎2

𝑥2 − 𝑎2 2

𝐸𝑥 = 𝑘𝑒𝑞4𝑎𝑥

𝑥2 − 𝑎2 2

When 𝑥 is much, much greater than 𝑎, we find:

𝐸𝑥 ≈ 4𝑘𝑒𝑞𝑎

𝑥3

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 12

23.04sec.

Problem 23.22

Consider the electric dipole shown in the following figure. Show that the electric field at a distant point on the +𝑥 axis is 𝐸𝑥 ≈ 4𝑘𝑒𝑞𝑎

𝑥3.

The Electric Field

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 13

23.04The Electric Fieldsec.

Problem 23.52

Three point charges are aligned along the 𝑥 axis as shown in the following figure. Find the electric field at (a) the position 2.00, 0 and (b) the position0, 2.00 .

𝐸1 = 𝑘𝑒𝑞1

𝑟12

𝐸1 = 8.99 × 109 ×

−4.00 × 10−9

2.50 2

𝐸1 = 5.75 ΤN C

𝐸2 = 𝑘𝑒𝑞2

𝑟22

𝐸2 = 8.99 × 109 ×

5.0 × 10−9

2.00 2

𝐸2 = 11.24 ΤN C

(a)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 14

23.04The Electric Fieldsec.

Problem 23.52

Three point charges are aligned along the 𝑥 axis as shown in the following figure. Find the electric field at (a) the position 2.00, 0 and (b) the position0, 2.00 .

𝐸3 = 𝑘𝑒𝑞3

𝑟32

𝐸3 = 8.99 × 109 ×

3.00 × 10−9

1.20 2

𝐸3 = 18.73 ΤN C

𝐸 = −𝐸1 + 𝐸2 + 𝐸3

𝐸 = −5.75 + 11.24 + 18.73

𝐸 = 24.22 ΤN C in +𝑥 direction.

(a)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 15

23.04The Electric Fieldsec.

Problem 23.52

Three point charges are aligned along the 𝑥 axis as shown in the following figure. Find the electric field at (a) the position 2.00, 0 and (b) the position0, 2.00 .

𝐸1 = 𝑘𝑒𝑞1

𝑟12

𝐸1 = 8.99 × 109 ×

−4.0 × 10−9

2.06 2

𝐸1 = 8.47 ΤN C

𝐸2 = 𝑘𝑒𝑞2

𝑟22

𝐸2 = 8.99 × 109 ×

5.0 × 10−9

2.00 2

𝐸2 = 11.24 ΤN C

(b)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 16

23.04The Electric Fieldsec.

Problem 23.52

Three point charges are aligned along the 𝑥 axis as shown in the following figure. Find the electric field at (a) the position 2.00, 0 and (b) the position0, 2.00 .

𝐸3 = 𝑘𝑒𝑞3

𝑟32

𝐸3 = 8.99 × 109 ×

3.0 × 10−9

2.15 2

𝐸3 = 5.83 ΤN C

(b)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 17

23.04The Electric Fieldsec.

Problem 23.52

𝐸𝑥 = 𝐸1𝑥 + 𝐸2𝑥 +𝐸3𝑥

𝐸𝑥 = 𝐸1 cos(𝜃1) + 𝐸2 cos(𝜃2) + 𝐸3 cos(𝜃3)

𝐸𝑥 = − 8.47 ×0.50

2.06+ 0 − 5.83 ×

0.80

2.15

𝐸𝑥 = −4.23 ΤN C

𝐸𝑦 = 𝐸1𝑦 + 𝐸2𝑦 +𝐸2𝑦

𝐸𝑦 = 𝐸1 sin(𝜃1) + 𝐸2 sin(𝜃2) + 𝐸3 sin(𝜃3)

𝐸𝑦 = − 8.47 ×2.00

2.06+ 11.24 + 5.83 ×

2.00

2.15

𝐸𝑦 = 8.44 ΤN C

Three point charges are aligned along the 𝑥 axis as shown in the following figure. Find the electric field at (a) the position 2.00, 0 and (b) the position0, 2.00 .

(b)

15/09/2020 Phys 104 - Ch. 23/IIII - lecture 4 - Dr. Alismail 18

23.04The Electric Fieldsec.

Problem 23.52

𝐸 = 𝐸𝑥2 + 𝐸𝑦

2

𝐸 = −4.23 2 + 8.44 2

𝐸 = 9.44 ΤN C

𝜑 = tan−1𝐸𝑦

𝐸𝑥

𝜑 = tan−18.44

−4.23

𝜑 = 116.62° !!

or

𝐸 = −4.23 Ƹ𝑖 + 8.44 Ƹ𝑗 ΤN C

Three point charges are aligned along the 𝑥 axis as shown in the following figure. Find the electric field at (a) the position 2.00, 0 and (b) the position0, 2.00 .

(b)

15/09/2020 Phys 104 - Ch. 23/I - lecture 1 - Dr. Alismail 19

23.06Electric Field Linessec.

Electric field lines penetrating two

surfaces. The magnitude of the field is

greater on surface A than on surfaceB.

The electric field lines for a point charge. (a) For a positive point charge, the lines are

directed radially outward. (b) For a negative point charge, the lines are directed radially

inward. Note that the figures show only those field lines that lie in the plane of the page. (c)

The dark areas are small pieces of thread suspended in oil, which align with the electric

field produced by a small charged conductor at the center.

➢ The electric field vector E is tangent to the electric field line at each point. The line has a direction, indicated by an arrowhead, that is the same as

that of the electric field vector.

➢ The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region.

Thus, the field lines are close together where the electric field is strong and far apart where the field is weak.

15/09/2020 Phys 104 - Ch. 23/I - lecture 1 - Dr. Alismail 20

23.06Electric Field Linessec.

(a) The electric field lines for two point charges of

equal magnitude and opposite sign (an electric

dipole). The number of lines leaving the positive

charge equals the number terminating at the

negative charge. (b) The dark lines are small pieces

of thread suspended in oil, which align with the

electric field of a dipole.

(a) The electric field lines for two positive point

charges. (b) Pieces of thread suspended in oil,

which align with the electric field created by two

equal-magnitude positive charges.

➢ The lines must begin on a positive charge and terminate on a negative charge. In the case of an excess of one type of charge, some lines will begin

or end infinitely far away.

➢ The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.

➢ No two field lines can cross.

The electric field lines for

a point charge +2𝑞 and asecond point charge −𝑞.Note that two lines leave

+ 2𝑞 for every one thatterminates on −𝑞.