FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
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Chapter 16, Problem 56QAP
To determine

(a)

The electric field at a distance x=0.06 m

Expert Solution
Check Mark

Answer to Problem 56QAP

Net electric field at ( x=0.06 m ) is 6.87×107N/C

Explanation of Solution

Given:

Charge 1, q1=+5×106C

Charge 2, q2=10×106C

Distance of the point from charge 1, r1=0.06 m

Distance of the point from charge 2, r2=0.04 m

  FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 16, Problem 56QAP , additional homework tip  1

Formula used:

Electric field at a distance x from the charge q.

  Ex=k|q|r2Where,r=distanceformthechargeqconstant,k=8.99×109Nm2C2

Calculation:

  E1x=k| q 1|r12=( 8.99× 10 9 N m 2 C 2 )|5.00× 10 6C| (0.060m)2=1.249×107NCE2x=k| q 2|r22=( 8.99× 10 9 N m 2 C 2 )|10.0× 10 6C| (0.040)2=5.619×107NC

Net electric field is the sum of the electric field due to both charges.

  Ex=E1x+E2x=1.249×107NC+5.619×107NC=6.87×107NC

Conclusion:

Net electric field at ( x=0.06 m ) is 6.87×107N/C

To determine

(b)

The point on the x -axis at which the electric field is zero.

Expert Solution
Check Mark

Answer to Problem 56QAP

The point where E = 0 is at distance x=24.1 cm.

Explanation of Solution

Given:

Charge 1, q1=+5×106C

Charge 2, q2=10×106C

Distance of the point from charge (1) is x

Distance of the point from charge (2) is (0.1x)

  FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 16, Problem 56QAP , additional homework tip  2

Formula used:

The electric field at a distance x from the charge q.

  Ex=k|q|r2Where,r=distanceformthechargeqconstant,k=8.99×109Nm2C2

Calculation:

  E=E1+E2=( k q 1 x 2 )+( k q 2 (x0.100) 2 )=0q1x2+q2 (x0.100)2=05.00× 10 6x2+10.00× 10 6 (x0.100)2=05x2+x0.05=0x=1± 1 2 4(5)(0.05)2(5)x=0.241m,+0.0414m

The point where E = 0 must lie to the left of both charges, where it is closer to the smaller charge

Conclusion:

The point where E = 0 is at distance x=24.1 cm.

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Chapter 16 Solutions

FlipIt for College Physics (Algebra Version - Six Months Access)

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