(a) The expression for electric field at point X .

Question

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Answer 1

Question

Chapter 16, Problem 80QAP

To determine

(a)

The expression for electric field at point X.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point X is, Ex=9600.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2

To determine

(b)

The expression for electric field at point Y.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Y is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Y is given by,
Ey=kqAryA2+kqBryB2+kqCryC2

To determine

(c)

The expression for electric field at point Z.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Z is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Z is given by,
Ez=kqArzA2+kqBrzB2+kqCrzC2

To determine

(d)

The electric field at point X, Y and Z.

Expert Solution

Answer to Problem 80QAP

The electric field at point X, Y and Z are, Ex=9600.56N/C, Ey=−11568.7N/C and Ez=−9240.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2rA=rB=rC=r=34×s=34×0.1=0.04330mEx=kqArA2+kqBrB2+kqCrC2=9× 109 ( 0.04330 )2(20.0×10−9−8.0×10−9−10.0×10−9)Ex=9600.56N/C

The electric field at point Y is given by,
Ey=kqAr yA2+kqBr yB2+kqCr yC2ryA=ryB=s2=5.0cm=0.5mryC=32×s=34×0.1=0.0866mEy=kqAr yA2+kqBr yB2+kqCr yC2Ey=9× 109 ( 0.5 )2(20.0×10−9−8.0×10−9)−9× 109×10.0× 10 −9 ( 0.0866 )2Ey=−11568.7N/C

The electric field at point Z is given by,
Ez=kqAr zA2+kqBr zB2+kqCr zC2rzA=rzC=s2=5.0cm=0.5mrzB=32×s=0.0866mEz=kqAr zA2+kqBr zB2+kqCr zC2Ez=9× 109 ( 0.5 )2(20.0×10−9−10.0×10−9)−9× 109×8.0× 10 −9 ( 0.0866 )2Ez=−9240.56N/C

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Answer 2

Question

Chapter 16, Problem 80QAP

To determine

(a)

The expression for electric field at point X.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point X is, Ex=9600.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2

To determine

(b)

The expression for electric field at point Y.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Y is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Y is given by,
Ey=kqAryA2+kqBryB2+kqCryC2

To determine

(c)

The expression for electric field at point Z.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Z is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Z is given by,
Ez=kqArzA2+kqBrzB2+kqCrzC2

To determine

(d)

The electric field at point X, Y and Z.

Expert Solution

Answer to Problem 80QAP

The electric field at point X, Y and Z are, Ex=9600.56N/C, Ey=−11568.7N/C and Ez=−9240.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2rA=rB=rC=r=34×s=34×0.1=0.04330mEx=kqArA2+kqBrB2+kqCrC2=9× 109 ( 0.04330 )2(20.0×10−9−8.0×10−9−10.0×10−9)Ex=9600.56N/C

The electric field at point Y is given by,
Ey=kqAr yA2+kqBr yB2+kqCr yC2ryA=ryB=s2=5.0cm=0.5mryC=32×s=34×0.1=0.0866mEy=kqAr yA2+kqBr yB2+kqCr yC2Ey=9× 109 ( 0.5 )2(20.0×10−9−8.0×10−9)−9× 109×10.0× 10 −9 ( 0.0866 )2Ey=−11568.7N/C

The electric field at point Z is given by,
Ez=kqAr zA2+kqBr zB2+kqCr zC2rzA=rzC=s2=5.0cm=0.5mrzB=32×s=0.0866mEz=kqAr zA2+kqBr zB2+kqCr zC2Ez=9× 109 ( 0.5 )2(20.0×10−9−10.0×10−9)−9× 109×8.0× 10 −9 ( 0.0866 )2Ez=−9240.56N/C

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Answer 3

Question

Chapter 16, Problem 80QAP

To determine

(a)

The expression for electric field at point X.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point X is, Ex=9600.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2

To determine

(b)

The expression for electric field at point Y.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Y is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Y is given by,
Ey=kqAryA2+kqBryB2+kqCryC2

To determine

(c)

The expression for electric field at point Z.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Z is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Z is given by,
Ez=kqArzA2+kqBrzB2+kqCrzC2

To determine

(d)

The electric field at point X, Y and Z.

Expert Solution

Answer to Problem 80QAP

The electric field at point X, Y and Z are, Ex=9600.56N/C, Ey=−11568.7N/C and Ez=−9240.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2rA=rB=rC=r=34×s=34×0.1=0.04330mEx=kqArA2+kqBrB2+kqCrC2=9× 109 ( 0.04330 )2(20.0×10−9−8.0×10−9−10.0×10−9)Ex=9600.56N/C

The electric field at point Y is given by,
Ey=kqAr yA2+kqBr yB2+kqCr yC2ryA=ryB=s2=5.0cm=0.5mryC=32×s=34×0.1=0.0866mEy=kqAr yA2+kqBr yB2+kqCr yC2Ey=9× 109 ( 0.5 )2(20.0×10−9−8.0×10−9)−9× 109×10.0× 10 −9 ( 0.0866 )2Ey=−11568.7N/C

The electric field at point Z is given by,
Ez=kqAr zA2+kqBr zB2+kqCr zC2rzA=rzC=s2=5.0cm=0.5mrzB=32×s=0.0866mEz=kqAr zA2+kqBr zB2+kqCr zC2Ez=9× 109 ( 0.5 )2(20.0×10−9−10.0×10−9)−9× 109×8.0× 10 −9 ( 0.0866 )2Ez=−9240.56N/C

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Answer 4

Question

Chapter 16, Problem 80QAP

To determine

(a)

The expression for electric field at point X.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point X is, Ex=9600.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2

To determine

(b)

The expression for electric field at point Y.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Y is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Y is given by,
Ey=kqAryA2+kqBryB2+kqCryC2

To determine

(c)

The expression for electric field at point Z.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Z is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Z is given by,
Ez=kqArzA2+kqBrzB2+kqCrzC2

To determine

(d)

The electric field at point X, Y and Z.

Expert Solution

Answer to Problem 80QAP

The electric field at point X, Y and Z are, Ex=9600.56N/C, Ey=−11568.7N/C and Ez=−9240.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2rA=rB=rC=r=34×s=34×0.1=0.04330mEx=kqArA2+kqBrB2+kqCrC2=9× 109 ( 0.04330 )2(20.0×10−9−8.0×10−9−10.0×10−9)Ex=9600.56N/C

The electric field at point Y is given by,
Ey=kqAr yA2+kqBr yB2+kqCr yC2ryA=ryB=s2=5.0cm=0.5mryC=32×s=34×0.1=0.0866mEy=kqAr yA2+kqBr yB2+kqCr yC2Ey=9× 109 ( 0.5 )2(20.0×10−9−8.0×10−9)−9× 109×10.0× 10 −9 ( 0.0866 )2Ey=−11568.7N/C

The electric field at point Z is given by,
Ez=kqAr zA2+kqBr zB2+kqCr zC2rzA=rzC=s2=5.0cm=0.5mrzB=32×s=0.0866mEz=kqAr zA2+kqBr zB2+kqCr zC2Ez=9× 109 ( 0.5 )2(20.0×10−9−10.0×10−9)−9× 109×8.0× 10 −9 ( 0.0866 )2Ez=−9240.56N/C

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Answer 5

Chapter 16, Problem 80QAP

To determine

(a)

The expression for electric field at point X.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point X is, Ex=9600.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2

To determine

(b)

The expression for electric field at point Y.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Y is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Y is given by,
Ey=kqAryA2+kqBryB2+kqCryC2

To determine

(c)

The expression for electric field at point Z.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Z is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Z is given by,
Ez=kqArzA2+kqBrzB2+kqCrzC2

To determine

(d)

The electric field at point X, Y and Z.

Expert Solution

Answer to Problem 80QAP

The electric field at point X, Y and Z are, Ex=9600.56N/C, Ey=−11568.7N/C and Ez=−9240.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2rA=rB=rC=r=34×s=34×0.1=0.04330mEx=kqArA2+kqBrB2+kqCrC2=9× 109 ( 0.04330 )2(20.0×10−9−8.0×10−9−10.0×10−9)Ex=9600.56N/C

The electric field at point Y is given by,
Ey=kqAr yA2+kqBr yB2+kqCr yC2ryA=ryB=s2=5.0cm=0.5mryC=32×s=34×0.1=0.0866mEy=kqAr yA2+kqBr yB2+kqCr yC2Ey=9× 109 ( 0.5 )2(20.0×10−9−8.0×10−9)−9× 109×10.0× 10 −9 ( 0.0866 )2Ey=−11568.7N/C

The electric field at point Z is given by,
Ez=kqAr zA2+kqBr zB2+kqCr zC2rzA=rzC=s2=5.0cm=0.5mrzB=32×s=0.0866mEz=kqAr zA2+kqBr zB2+kqCr zC2Ez=9× 109 ( 0.5 )2(20.0×10−9−10.0×10−9)−9× 109×8.0× 10 −9 ( 0.0866 )2Ez=−9240.56N/C

Answer 6

Chapter 16, Problem 80QAP

To determine

(a)

The expression for electric field at point X.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point X is, Ex=9600.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2

Answer 7

Chapter 16, Problem 80QAP

To determine

(a)

The expression for electric field at point X.

Answer 8

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point X is, Ex=9600.56N/C

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2

Answer 13

To determine

(b)

The expression for electric field at point Y.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Y is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Y is given by,
Ey=kqAryA2+kqBryB2+kqCryC2

Answer 14

To determine

(b)

The expression for electric field at point Y.

Answer 15

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Y is, Ey=−11568.7N/C

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Y is given by,
Ey=kqAryA2+kqBryB2+kqCryC2

Answer 20

To determine

(c)

The expression for electric field at point Z.

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Z is, Ey=−11568.7N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Z is given by,
Ez=kqArzA2+kqBrzB2+kqCrzC2

Answer 21

To determine

(c)

The expression for electric field at point Z.

Answer 22

Expert Solution

Answer to Problem 80QAP

The expression for electric field at point Z is, Ey=−11568.7N/C

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

The electric field at point Z is given by,
Ez=kqArzA2+kqBrzB2+kqCrzC2

Answer 27

To determine

(d)

The electric field at point X, Y and Z.

Expert Solution

Answer to Problem 80QAP

The electric field at point X, Y and Z are, Ex=9600.56N/C, Ey=−11568.7N/C and Ez=−9240.56N/C

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2rA=rB=rC=r=34×s=34×0.1=0.04330mEx=kqArA2+kqBrB2+kqCrC2=9× 109 ( 0.04330 )2(20.0×10−9−8.0×10−9−10.0×10−9)Ex=9600.56N/C

The electric field at point Y is given by,
Ey=kqAr yA2+kqBr yB2+kqCr yC2ryA=ryB=s2=5.0cm=0.5mryC=32×s=34×0.1=0.0866mEy=kqAr yA2+kqBr yB2+kqCr yC2Ey=9× 109 ( 0.5 )2(20.0×10−9−8.0×10−9)−9× 109×10.0× 10 −9 ( 0.0866 )2Ey=−11568.7N/C

The electric field at point Z is given by,
Ez=kqAr zA2+kqBr zB2+kqCr zC2rzA=rzC=s2=5.0cm=0.5mrzB=32×s=0.0866mEz=kqAr zA2+kqBr zB2+kqCr zC2Ez=9× 109 ( 0.5 )2(20.0×10−9−10.0×10−9)−9× 109×8.0× 10 −9 ( 0.0866 )2Ez=−9240.56N/C

Answer 28

To determine

(d)

The electric field at point X, Y and Z.

Answer 29

Expert Solution

Answer to Problem 80QAP

The electric field at point X, Y and Z are, Ex=9600.56N/C, Ey=−11568.7N/C and Ez=−9240.56N/C

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given:

Charge, qA=+20.0nC=20.0×10−9C

Charge, qB=−8.00nC=−8.00×10-9C

Charge, qC=−10.0nC=−10.0×10−9C

Distance, s=10.0cm=0.10m

Formula used:

The electric field is given by,
E=kqr2

Where,
E =Electric field
q = Electric charges
r =Distance from charge

Calculation:

The electric field at point X is given by,
Ex=kqArA2+kqBrB2+kqCrC2rA=rB=rC=r=34×s=34×0.1=0.04330mEx=kqArA2+kqBrB2+kqCrC2=9× 109 ( 0.04330 )2(20.0×10−9−8.0×10−9−10.0×10−9)Ex=9600.56N/C

The electric field at point Y is given by,
Ey=kqAr yA2+kqBr yB2+kqCr yC2ryA=ryB=s2=5.0cm=0.5mryC=32×s=34×0.1=0.0866mEy=kqAr yA2+kqBr yB2+kqCr yC2Ey=9× 109 ( 0.5 )2(20.0×10−9−8.0×10−9)−9× 109×10.0× 10 −9 ( 0.0866 )2Ey=−11568.7N/C

The electric field at point Z is given by,
Ez=kqAr zA2+kqBr zB2+kqCr zC2rzA=rzC=s2=5.0cm=0.5mrzB=32×s=0.0866mEz=kqAr zA2+kqBr zB2+kqCr zC2Ez=9× 109 ( 0.5 )2(20.0×10−9−10.0×10−9)−9× 109×8.0× 10 −9 ( 0.0866 )2Ez=−9240.56N/C