General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Question
Chapter 16, Problem 74E

(a)

To determine

The magnitude and direction of the field due to the charges.

(a)

Expert Solution
Check Mark

Answer to Problem 74E

The magnitude of net electric field above the surface of the earth is (45000N/C)_ and is directed downward towards the ground_.

Explanation of Solution

“The electric force per unit charge is defined as the electric field”.

The expression for the electric field due to a point charge is as follows:

  E=kQr2r^        (I)

Here, E is the electric field, Q is the charge, k is the Coulomb’s constant, and r is the distance.

Analogous to equation (I), the electric field at the surface of the earth just below the cloud due to the charge +50C can be written as,

  E1=kQ1r12j^        (II)

Here, E1 is the electric field due to the charge +50C and r1 is the distance between the charge 1 and the ground.

Analogous to equation (I), the electric field at the surface of the earth just below the cloud due to the charge 20C can be written as,

  E2=kQ2r22(j^)        (III)

Here, E2 is the electric field due to the charge 20C and r2 is the distance between the charge 2 and the ground.

Write the expression for the net electric field.

  Enet=E1E2        (IV)

Here, Enet is the net electric field.

Use equation (II) and (III) in equation (IV).

  Enet=kQ1r12j^kQ2r22(j^)

Conclusion:

Substitute 9×109Nm2/C2 for k, +50C for Q1, 10km for r1, 20C for Q2, and 2km for r2.

  Enet=(9×109Nm2/C2)(+50C)(10km)2j^(9×109Nm2/C2)(20C)(2km)2(j^)=(9×109Nm2/C2)(+50C)(10km×103m1km)2j^(9×109Nm2/C2)(20C)(2km×103m1km)2(j^)==(9×109Nm2/C2)(+50C)(10km×103m1km)2j^(9×109Nm2/C2)(20C)(2km×103m1km)2j^=(4500N/C)j^(45000N/C)j^=(45000N/C)(j^)

The direction of the electric field due to both the charges at the surface of the earth is downward.

Therefore, the magnitude of net electric field above the surface of the earth is (45000N/C) and is directed downward towards the ground.

(b)

To determine

The locations and values of the image charges corresponding to the +50C and 20C charges.

(b)

Expert Solution
Check Mark

Answer to Problem 74E

The locations and values of the image charges corresponding to the +50C and 20C charges is shown in figure 1.

Explanation of Solution

The image charge for +50C is 50C and the image charge for 20C is +20C and they are located at the same distance below the surface as of the same distance they have located above the surface.

Conclusion:

The image charge for +50C is 50C and the image charge for 20C is +20C is shown below.

General Physics, 2nd Edition, Chapter 16, Problem 74E

(c)

To determine

The total electric field due to all the charges at the surface of earth just below the cloud.

(c)

Expert Solution
Check Mark

Answer to Problem 74E

The total electric field due to the charges at the surface of the earth just below the cloud is (20250N/C)j^_.

Explanation of Solution

“The electric force per unit charge is defined as the electric field”.

The expression for the electric field due to a point charge is as follows:

  E=kQr2r^

Here, E is the electric field, Q is the charge, k is the Coulomb’s constant, and r is the distance.

Write the expression for the electric field due to the charges 20C and 20C is,

  E1'=kQ+20cd12(j)+kQ20cd12(j)        (V)

Here, E1' is the electric field due to the charges +20C and 20C and d1 is the distance between the charges between +20C and 20C.

Write the expression for the electric field due to the charges 50C and 50C is,

  E2'=kQ+50cd22(j)+kQ50cd22(j)        (VI)

Here, E2' is the electric field due to the charges +50C and 50C and d2 is the distance between the charges between +20C and 20C.

Write the expression for the total electric field.

  Etotal=E1'+E2'        (VII)

Here, Etotal is the net electric field.

Use equation (V) and (VI) in equation (VII).

  Etotal=kQ+20cd12(j)+kQ20cd12(j)+kQ+50cd22(j)+kQ50cd22(j)

Conclusion:

Substitute 9×109Nm2/C2 for k, 50C for Q+50C, 50C for Q50C 20km for d1, +20C for Q+20C, 20C for Q20C and 4km for d2.

  Etotal=(9×109Nm2/C2)(20C)(4km)2(j^)+(9×109Nm2/C2)(20C)(4km)2(j^)+(9×109Nm2/C2)(50C)(20km)2(j^)+(9×109Nm2/C2)(50C)(20km)2(j^)=(9×109Nm2/C2)(20C)(4km(103m1km))2(j^)+(9×109Nm2/C2)(20C)(4km(103m1km))2(j^)+(9×109Nm2/C2)(50C)(20km(103m1km))2(j^)+(9×109Nm2/C2)(50C)(20km(103m1km))2(j^)=(22500N/C)j^+(2250N/C)j^=(20250N/C)j^

Therefore, the total electric field due to the charges at the surface of the earth just below the cloud is (20250N/C)j^.

(d)

To determine

Whether the lightning from this cloud strike the ground or not.

(d)

Expert Solution
Check Mark

Answer to Problem 74E

No, the lightning will not strike the ground.

Explanation of Solution

The total electric field due to all the charges is less than the electric field generated by the air as a conductor. Since the electric field due to all the charges is less than the electric field generated by the air as a conductor, the cloud lightning does not strike the ground.

Conclusion:

Since the electric field due to all the charges is less than the electric field generated by the air as a conductor, the cloud lightning does not strike the ground.

Therefore, the lightning from the cloud does not strike the ground.

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Chapter 16 Solutions

General Physics, 2nd Edition

Ch. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - Prob. 31ECh. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - Prob. 49ECh. 16 - Prob. 50ECh. 16 - Prob. 51ECh. 16 - Prob. 52ECh. 16 - Prob. 53ECh. 16 - Prob. 54ECh. 16 - Prob. 55ECh. 16 - Prob. 56ECh. 16 - Prob. 57ECh. 16 - Prob. 58ECh. 16 - Prob. 59ECh. 16 - Prob. 60ECh. 16 - Prob. 61ECh. 16 - Prob. 62ECh. 16 - Prob. 63ECh. 16 - Prob. 64ECh. 16 - Prob. 65ECh. 16 - Prob. 66ECh. 16 - Prob. 67ECh. 16 - Prob. 68ECh. 16 - Prob. 69ECh. 16 - Prob. 70ECh. 16 - Prob. 72ECh. 16 - Prob. 73ECh. 16 - Prob. 74ECh. 16 - Prob. 75ECh. 16 - Prob. 76ECh. 16 - Prob. 78ECh. 16 - Prob. 81ECh. 16 - Prob. 82ECh. 16 - Prob. 83ECh. 16 - Prob. 84ECh. 16 - Prob. 85ECh. 16 - Prob. 86ECh. 16 - Prob. 87ECh. 16 - Prob. 88ECh. 16 - Prob. 89ECh. 16 - Prob. 90ECh. 16 - Prob. 91ECh. 16 - Prob. 92ECh. 16 - Prob. 93ECh. 16 - Prob. 94ECh. 16 - Prob. 95ECh. 16 - Prob. 96ECh. 16 - Prob. 97ECh. 16 - Prob. 98ECh. 16 - Prob. 99ECh. 16 - Prob. 100ECh. 16 - Prob. 101ECh. 16 - Prob. 102ECh. 16 - Prob. 103ECh. 16 - Prob. 104ECh. 16 - Prob. 105ECh. 16 - Prob. 106ECh. 16 - Prob. 107ECh. 16 - Prob. 108E
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