Electromagnetic theory Drill Problrm Sol, Exercises of Engineering

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E E 0 8. S O L U T I O N S

DRILL PROBLEMS : CHAPTER 2

D2.

( a) RAB = (5+6) ax + (8-4) ay + (-2-7) az = 11 ax + 4 ay - 9 az

( b) RAB = 112 + 4^2 + 9^2 = 14.76 m

( c) FBA = −20 × 10−^6 50 × 10−^6 4 𝜋 10 − 9 36 𝜋 (14.76^2 )

(− 11 𝑎𝑥 − 4 𝑎𝑦 + 9𝑎𝑧 ) 14.76 = 30.78𝑎𝑥^ + 11.195𝑎𝑦^ −^ 25.18𝑎𝑧^ mN

( d) FBA = −20 × 10

− (^6) 50 × 10− 6 4 𝜋 ×8.854×10−^12 (14.76^2 ) 𝑎𝑏𝑎^ =^ −0.^

(− 11 𝑎𝑥 − 4 𝑎𝑦 + 9𝑎𝑧 ) 14.76 = 30.74𝑎𝑥^ + 11.18𝑎𝑦^ −^ 25.15𝑎𝑧^ mN

D2.

(a) 𝒓 − 𝒓𝑨 = − 25 𝑎𝑥 + 30𝑎𝑦 − 15 𝑎𝑧 , |𝒓 − 𝒓𝑨| = 41.

𝒓 − 𝒓𝑩 = 10𝑎𝑥 − 8 𝑎𝑦 − 12 𝑎𝑧 , |𝒓 − 𝒓𝑩| = 17. 𝑬𝑨 = −1. (− 25 𝑎𝑥 +30𝑎𝑦 − 15 𝑎𝑧 ) 41.43 = 9480𝑎𝑥^ −^11300 𝑎𝑦^ + 5600𝑎𝑧 𝑬𝑩 = 14. (10𝑎𝑥 − 8 𝑎𝑦 − 12 𝑎𝑧 ) 17.54 = 83300𝑎𝑥^ −^66600 𝑎𝑦^ −^99900 𝑎𝑧 𝑬𝑻 = 𝑬𝑨 + 𝑬𝑩 = 92.48𝑎𝑥 − 77.9𝑎𝑦 − 94.3𝑎𝑧^ 𝑘𝑉𝑚

(b) 𝒓 − 𝒓𝑨 = − 10 𝑎𝑥 + 50𝑎𝑦 + 35𝑎𝑧 , |𝒓 − 𝒓𝑨| = 61.

𝒓 − 𝒓𝑩 = 25𝑎𝑥 + 12𝑎𝑦 + 38𝑎𝑧 , |𝒓 − 𝒓𝑩| = 47. 𝑬𝑨 = − 7050 (−^10 𝑎𝑥^ +5061.84𝑎𝑦 +35𝑎𝑧^ )= 1140𝑎𝑥 − 5700 𝑎𝑦 − 3990 𝑎𝑧 𝑬𝑩 = 20300 (25𝑎𝑥 +12𝑎𝑦 +38𝑎𝑧 ) 47.04 = 10700𝑎𝑥^ + 5180𝑎𝑦^ + 16400𝑎𝑧 𝑬𝑻 = 𝑬𝑨 + 𝑬𝑩 = 11.84𝑎𝑥 − 0.52𝑎𝑦 + 12.41𝑎𝑧^ 𝑘𝑉𝑚

D2.

(a ) Sum = 2 + 0 + 25 + 0 + 172 + 0 = 2.

( b) Sum = (^) 11.181.1 + (^) 22.621.01 + 1.00146.87 + 1.000189.44 = 0.

Made by Zaeem Ahmad Varaich Please inform if you find any error in any solution!

E E 0 8. S O L U T I O N S

D2.

(a ) 𝑄 = (^) 𝑣𝑜𝑙 𝜌𝑣𝑑𝑣 = 1 𝑥^3 𝑦 3 𝑧^3 𝑑𝑥𝑑𝑦𝑑𝑧^ +

−0. −0.

−0. −0.

−0. −0.

1 𝑥^3 𝑦 3 𝑧^3 𝑑𝑥𝑑𝑦𝑑𝑧

= − (^8) 𝑥 2 1 −^ − 02 0.1^ 𝑦^2 −^ − 02 0.1^ 𝑧^2 −^ − 02 0.^

0.10.2^ 𝑦^2 0.10.2^ 𝑧^2 0.10.^

= 8×(0.03)^1 − 8×(0.03)^1 = 0

(b ) (^) 0 𝜋 0 0.1 24 𝜌^3 𝑧^2 sin 0.6𝜑 𝑑𝑧𝑑𝜌𝑑𝜑= (− cos 0.6 0.6 𝜑) 0

𝜋 (𝜌

4 4 )^0

(𝑧^

3 3 )^2

4 = 1.018 𝑚𝐶

(c ) (^) 0 ∞ 02 𝜋 02 𝜋𝑒−^2 𝑟^ sin 𝜃 𝑑𝜑𝑑𝜃𝑑𝑟 = (− 𝑒^

− 2 𝑟 2 )^0

∞ (cos 𝜃) 02 𝜋^ (𝜑) 02 𝜋^ = −6.28 𝐶

D2.

(a) E = 2 × 5×

− 9 2 𝜋𝜀𝑜 (4) 𝒂𝒛^ = 44.^

𝑉 𝑚

(b) Ex = 5×

− 9 2 𝜋𝜀𝑜 (5)

(3𝒂𝒚+4𝒂𝒛) 5 = 10.788𝒂𝒚^ + 14.384𝒂𝒛

𝑉 𝑚 ,^ Ey^ =^

5×10−^9 2 𝜋𝜀𝑜 (4) 𝒂𝒛^ = 22.4775^ 𝒂𝒛

𝑉 𝑚 E = Ex + Ey = 10.788𝒂𝒚 + 36.86𝒂𝒛 𝑉 𝑚

D2.

i) Electric field due to 3 nC/m^2 : E 1 = 3×

− 9 2 𝜀𝑜^ 𝒂𝑵=^ 169.5^ 𝒂𝒛 ii) Electric field due to 6 nC/m^2 : E 2 = 6×

− 9 2 𝜀𝑜^ 𝒂𝑵=^ 338.8^ 𝒂𝒛 iii) Electric field due to -8 nC/m^2 : E 3 = −8×

− 9 2 𝜀𝑜^ 𝒂𝑵=^ −451.76^ 𝒂𝒛

According to the direction of point relative to normal:

(a ) E = - E 1 - E 2 - E 3 = −56.6 𝒂𝒛𝑚^ 𝑉

(a) E = E 1 - E 2 - E 3 = 283 𝒂𝒛𝑚^ 𝑉

(a) E = E 1 + E 2 - E 3 = 961 𝒂𝒛𝑚^ 𝑉

(a) E = E 1 + E 2 + E 3 = 56.6 𝒂𝒛𝑚^ 𝑉

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E E 0 8. S O L U T I O N S

DRILL PROBLEMS 3

D3.

(a) Evaluate the triple volume integral to find the total volume enclosed by the portion of sphere / surface and then just multiply it with the given charge to find the total change within it:

𝑟^2 𝑠𝑖𝑛𝜃 𝑑𝜃𝑑𝜙𝑑𝑟 ×

𝜋 2

0

𝜋 2

0

0

× 𝑞 = 7.5𝜇𝐶

(b) This surface encloses the whole charge q , so answer is 60 μC (c) Only the upper half of the flux lines pass through the plane at z = 26 cm, so D = 0.5 x 60 = 30 μC

D3.

(a) 𝐸 =

𝑘𝑄 𝑟 2

4 𝑎𝑥 − 6 𝑎𝑦 +12𝑎𝑧

16+36+144 = 0.72𝑎𝑥^ −^ 1.08𝑎𝑦^ + 2.16𝑎𝑧

𝑀𝑉 𝑚

𝑚^2

(b) 𝐸 =

20 2 𝜋𝜀𝑜.

− 3 𝑎𝑦 +6𝑎𝑧 45

𝑀𝑉 𝑚

𝑚^2

(c) 𝐸 =

120

2 𝜀𝑜^ 𝑎𝑧^ =^

60

𝜀𝑜^ 𝑎𝑧

𝜇 𝑉

𝑚 ,^ 𝑠𝑜^ 𝐷^ =^ 𝜀𝑜^ 𝐸^ = 60𝑎𝑧

𝜇𝐶 𝑚 2

D3.

(a) 𝐸 =

𝐷

𝜀𝑜^ = 33.88𝑟

𝑟 , so at P:^ 𝐸^ = 33.88(2)

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E E 0 8. S O L U T I O N S

(b) 𝑄 =

𝐷. 𝑑𝑠 = 𝑎^2 𝑠𝑖𝑛𝜃 𝑑𝜃𝑑𝜙𝑎𝑟 × 0.3𝑟^2 𝑎𝑟 =

𝜋 0

2 𝜋

0 24.3^ −𝑐𝑜𝑠𝜃|^0

2 𝜋 𝜋

0 𝑑𝜙^ =

2 𝜋

0. 208^ 𝑛𝐶

(c) On same steps: 𝑄 = 76.8 −𝑐𝑜𝑠𝜃| 0 𝜋^ 𝑑𝜙 = 964.608 μC

2 𝜋 0

D3.

(a) 𝑄 = 𝑄 1 + 𝑄 2 = 0.243 𝜇𝐶

(b) 𝑄 = 𝑙𝑒𝑛𝑔𝑡𝑕 × 𝜌𝐿 = 31.4 𝜇𝐶

(c) Area = 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑙𝑖𝑛𝑒 (𝑕𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒) × 𝑤𝑖𝑑𝑡𝑕 𝑎𝑐𝑟𝑜𝑠𝑠 𝑧 = 10.53 × 10

So, 𝑄 = 𝑎𝑟𝑒𝑎 × 𝜌𝐴 = 10.53 𝜇𝐶

D3.

(a) 𝐷 =

4 𝜋(0.005)^2 = 795.77^ 𝜇𝐶

(b) 𝑄 2 = 4𝜋 0.01 2 × 𝜌𝑠 2 = 2.51 𝜇𝐶,

0.25+2.

4 𝜋(0.015)^2 = 977^ 𝜇𝐶

(c) 𝑄 3 = 4𝜋 0.018 2 × 𝜌𝑠 3 = −2.44 𝜇𝐶,

0.25+2.51−2.

4 𝜋 (0.025)^2 = 40.74^ 𝜇𝐶

(d) 𝐷 =

0.25+2.51−2.44+4𝜋 0.03 2 ×𝜌𝑠

4 𝜋(0.035)^2 = 0 , so^ 𝜌𝑠^4 = -28.29^ 𝜇𝐶

D3.

(a) 𝜓 = 𝐷. 𝑑𝑠 = 16 𝑥^2 𝑦𝑧^3 𝑑𝑥𝑑𝑦 =

2 0

3

0

3 1

10

10

10 3

𝑦 = 3 𝑥

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E E 0 8. S O L U T I O N S

5

0

2

0

𝜑 2

5

0

2

0

5

0

𝜋

0

𝜑 2

| 0 𝜋^ × 𝑧 | 05 − 1.

𝜌^2

| 02 × 𝑧 | 05

= − 48 0 − 1 5 − 15

= 225

Solved by Aqeel Anwar ( aqeelanwar.co.cc ) Digitized by Zaeem ( ee08.net.tc )

Solved by Aqeel Anwar ( aqeelanwar.co.cc ) Digitized by Zaeem ( ee08.net.tc )

Solved by Aqeel Anwar ( aqeelanwar.co.cc ) Digitized by Zaeem ( ee08.net.tc )

Solved by Aqeel Anwar ( aqeelanwar.co.cc ) Digitized by Zaeem ( ee08.net.tc )

Solved by Aqeel Anwar ( aqeelanwar.co.cc ) Digitized by Zaeem ( ee08.net.tc )

Solved by Aqeel Anwar ( aqeelanwar.co.cc ) Digitized by Zaeem ( ee08.net.tc )

Solved by Aqeel Anwar ( aqeelanwar.co.cc ) Digitized by Zaeem ( ee08.net.tc )

Solved by Aqeel Anwar ( aqeelanwar.co.cc ) Digitized by Zaeem ( ee08.net.tc )

Solved by Zaeem. Please report if you find any mistake!

E E 0 8. S O L U T I O N S

DRILL PROBLEMS 5

D5.

(a) At P:

ܬ= 10ሺ 3 ሻ^2 ሺ 2 ܽሻߩെ 4 ሺ 3 ݋ܿሻݏ^2 ሺ 30 ܽሻ߮^ = 180ܽ ߩെ 9 ܽ ߮

(b) Using formula (2):

ܫ= න න 10 ߩ^3 ܽݖߩ. ݀ ߶݀ݖ

2

2 ߨ

0

2

2 ߨ

0

ݖ^2

D5.

(a) Using formula (2):

ܫ= െන න 106 ݖ1.5^ ܽ ݖ.߶ ݀ߩ ݀ߩ

20 ߤ

0

2 ߨ

0

20 ߤ

0

2 ߨ

0

= െ 106 ሺ0.1ሻ1.5^ ቆ

ߩ^2

ቇ| 020 ߤ ሺ߶ሻ| 02 ߨ^ = െ39.7ߤܣ

(b) Using formula (3):

2 × 10^6

3 (c) Same formula: