EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 21, Problem 76P

(a)

To determine

The magnitude and the direction of the force exerted.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Four point charges placed at corners of a square whose edge length is L.

Formula used:

Write the expression for the net force acting on F1 .

  F1=F2,1+F3,1+F4,1   ........ (1)

Here, F1 is the net force on charge q1 , F2,1 is the force on charge q1 due to the charge q2 , F3,1 is the charge on charge q1 due to the charge q3 , F4,1 is the force on charge q1 due o the charge q4 .

Write the expression for the F2,1 that is the force on charge q1 due to the charge q2 .

  F2,1=kq1q2r2,1r^2,1   ........ (2)

Here, k is the constant.

Write the expression for the F3,1 that is the force on charge q1 due to the charge q3 .

  F3,1=kq1q2(r3,1)2r^3,1   ........ (3)

Here, r3,1 is the distance between the first charge and the third charge.

The magnitude of charge q4 is three times that of the other charges that is:

  F4,1=3kq1q2(r4,1)2r^4,1   ........ (4)

Here, r4,1 is the distance between the fourth charge and the first charge.

Substitute q for q1 , q for q1 and L for r2,1 in equation (2).

  F2,1=k(q)qL3(Lj^)F2,1=kq2L2j^

Substitute q for q1 , q for q2 and L for r3,1 in equation (3).

  F3,1=kq2232L3(Li^Lj^)F3,1=kq2232L2(i^+j^)

Substitute q for q1 , q for q1 and L for r4,1 in equation (4).

  F4,1=k(q)qL3(Li^)F4,1=kq2L2(i^)

Calculation:

Substitute kq2L2j^ for F2,1 , kq2232L2(i^+j^) for F3,1 , kq2L2(i^) for F4,1 in equation (1).

  F1=kq2L2j^+kq2232L2(i^+j^)+kq2L2(i^)F1=kq2L2(1122)(i^+j^)

Conclusion:

Thus, the magnitude and the direction of the force is kq2L2(1122)(i^+j^) .

(b)

To determine

The magnitude of the electric field.

(b)

Expert Solution
Check Mark

Explanation of Solution

Formula used:

Write the expression for the electric field at EP .

  EP=E1+E2+E3+E4   ......... (5)

Here, EP is the electric field at the point where the electric field is calculated, E1 is the electric field due to charge q1 , E2 is the electric field due to charge q2 , E3 is the electric field due to charge q3 and E4 is the electric field due to charge q4 .

Write the expression for the electric field due to charge q1 .

  E1=kqr1,p2(L2j^)   ......... (6)

Here, r1.p is the distance between the first charge and point p and k is constant.

Write the expression for the electric field due to charge q2 .

  E2=k(q)r2,p2(L2j^)   ......... (7)

Here, r2.p is the distance between the second charge and point p and k is constant.

Write the expression for the electric field due to charge q3 .

  E3=kq3r3,p2(Li^L2j^)   ........ (8)

Here, r3.p is the distance between the third charge and point p and k is constant.

Write the expression for the electric field due to charge q3 .

  E4.p=k(q)r4,p2(Li^L2j^)   ........ (9)

Here, r4.p is the distance between the fourth charge and point p and k is constant.

Substitute L2 for r1,p in equation (6).

  E1=kq(L2)2(L2j^)E1=4kqL2j^

Substitute L2 for r2,p in equation (7).

  E2=k(q)(L2)2(L2j^)E2=4kqL2j^

Substitute L for r3,p in equation (8).

  E3=kq3532L2(i^12j^)

Substitute L for r4,p in equation (9).

  E4=kq4532L2(i^12j^)

Calculation:

Substitute 4kqL2j^ for E1 , 4kqL2j^ for E2 , kq3532L2(i^12j^) for E3 and kq4532L2(i^12j^) for E4 in equation (5).

  EP=4kqL2j^+4kqL2j^+kq3532L2(i^12j^)+kq4532L2(i^12j^)EP=8kqL2(1+525)j^

Conclusion:

Thus, the magnitude and the direction of the electric field is 8kqL2(1+525)j^ .

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Chapter 21 Solutions

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