College Physics (Instructor's)
College Physics (Instructor's)
11th Edition
ISBN: 9781305965317
Author: SERWAY
Publisher: CENGAGE L
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Chapter 15, Problem 24P

(a) Find the magnitude and direction of the electric field at the position of the 2.00 μC charge in Figure P13.13. (b) How would the electric field at that point be affected if the charge there were doubled? Would the magnitude of the electric force be affected?

(a)

Expert Solution
Check Mark
To determine
The magnitude and direction of net electric field on 2μC .

Answer to Problem 24P

Solution:

The magnitude of net electric field on 2μC is 2.19×105N/C .

The direction of net electric field on 2μC is 85.2ο below the x axis.

Explanation of Solution

The force diagram is given by,

College Physics (Instructor's), Chapter 15, Problem 24P

In the above diagram,

  • F1 is the force due to q1 on q2 .
  • F3 is the force due to q3 on q2 .

Formula to calculate the force due to q1 on q2 is,

F1=keq1q2a2      (I)

  • ke is the Coulomb constant.
  • a is the side of the triangle

Formula to calculate the force due to q3 on q2 is,

F3=keq3q2a2       (II)

Net force along the x direction is,

Fx=F3F1cos60ο       (III)

Net force along the y direction is,

Fy=F1sin60ο       (IV)

Formula to calculate the magnitude of net electric force is,

FR=Fx2+Fy2       (V)

Substitute Equations (III) and (IV) in (V).

FR=(keq3q2a2(keq1q2a2)cos60ο)2+(keq1q2a2sin60ο)2=(keq2a2)(q3q1cos60ο)2+(q1sin60ο)2

Formula to calculate the resultant electric field is,

ER=FRq2

Therefore,

E=(q2a2)(q3q1cos60ο)2+(q1sin60ο)2

Substitute 8.99×109N.m2/C2 for ke , 7.00μC for q1 , 2.00μC for q2 , 4.00μC for q3 and 0.500 m for a.

ER=[(8.99×109N.m2/C2)(0.500m)2]((4.00μC)(7.00μC)cos60ο)2+((7.00μC)sin60ο)2=[(8.99×109N.m2/C2)(0.500m)2]((4.00×106C)(7.00×106C)cos60ο)2+((7.00×106C)sin60ο)2=2.19×105N/C

Formula to calculate the direction is,

θ=tan1(FyFx)

From Equations (I), (II), (III) and (IV) of section 1,

θ=tan1(keq1q2a2sin60οkeq3q2a2(keq1q2a2)cos60ο)=tan1(q1sin60οq3q1cos60ο)

Substitute 7.00μC for q1 , 2.00μC for q2 and 4.00μC for q3 .

θ=tan1((7.00μC)sin60ο(4.00μC)(7.00μC)cos60ο)=85.2ο

Conclusion:

The magnitude of net electric field on 2μC is 0.438 N.

(b)

Expert Solution
Check Mark
To determine
The electric field if the charge is doubled.

Answer to Problem 24P

Solution: The electric field remains the same.

Explanation of Solution

When the charge is doubled, the electric force also gets doubled. Therefore, the electric field remains the same, since it is equal to the ration of electric force and charge.

Conclusion:

The electric field remains the same

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Chapter 15 Solutions

College Physics (Instructor's)

Ch. 15 - A glass object receives a positive charge of +3 nC...Ch. 15 - The fundamental charge is e = 1.60 1019 C....Ch. 15 - Each of the following statements is related to...Ch. 15 - Two uncharged, conducting spheres are separated by...Ch. 15 - Four concentric spheres S1, S2, S3, and S4 are...Ch. 15 - IF a suspended object A is attracted to a charged...Ch. 15 - Positive charge Q is located at the center of a...Ch. 15 - Consider point A in Figure CQ15.8 located an...Ch. 15 - A student stands on a thick piece of insulating...Ch. 15 - In fair weather, there is an electric field at the...Ch. 15 - A charged comb often attracts small bits of dry...Ch. 15 - Why should a ground wire be connected to the metal...Ch. 15 - There are great similarities between electric and...Ch. 15 - A spherical surface surrounds a point charge q....Ch. 15 - If more electric field lines leave a Gaussian...Ch. 15 - A student who grew up in a tropical country and is...Ch. 15 - What happens when a charged insulator is placed...Ch. 15 - A 7.50-nC charge is located 1.80 m from a 4.20-nC...Ch. 15 - A charged particle A exerts a force of 2.62 N to...Ch. 15 - Rocket observations show that dust particles in...Ch. 15 - A small sphere of mass m = 7.50 g and charge q1 =...Ch. 15 - The nucleus of 8Be, which consists of 4 protons...Ch. 15 - A molecule of DNA (deoxyribonucleic acid) is 2.17...Ch. 15 - Two uncharged spheres are separated by 2.00 in. 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(a)...Ch. 15 - Three equal positive charges are at the corners of...Ch. 15 - Refer 10 Figure 15.20. The charge lowered into the...Ch. 15 - The dome of a Van de Graaff generator receives a...Ch. 15 - If the electric field strength in air exceeds 3.0 ...Ch. 15 - In the Millikan oil-drop experiment illustrated in...Ch. 15 - A Van de Graaff generator is charged so that a...Ch. 15 - A uniform electric field of magnitude E = 435 N/C...Ch. 15 - An electric field of intensity 3.50 kN/C is...Ch. 15 - The electric field everywhere on the surface of a...Ch. 15 - Four closed surfaces, S1 through S4, together with...Ch. 15 - A charge q = +5.80 C is located at the center of a...Ch. 15 - Figure P15.49 shows a closed cylinder with...Ch. 15 - A charge of q = 2.00 109 G is spread evenly on a...Ch. 15 - A point charge q is located at the center of a...Ch. 15 - A charge of 1.70 102 C is at the center of a cube...Ch. 15 - Suppose the conducting spherical shell of Figure...Ch. 15 - A very large nonconducting plate lying in the...Ch. 15 - In deep spare, two spheres each of radius 5.00 m...Ch. 15 - A nonconducting, thin plane sheet of charge...Ch. 15 - Three point charges are aligned along the x-axis...Ch. 15 - A small plastic ball of mass m = 2.00 g is...Ch. 15 - A proton moving at v0 = 1.50 106 m/s enters the...Ch. 15 - The electrons in a particle beam each have a...Ch. 15 - A point charge +2Q is at the origin and a point...Ch. 15 - A 1.00-g cork ball having a positive charge of...Ch. 15 - Two 2.0-g spheres are suspended by 10.0-cm-long...Ch. 15 - a point charge of magnitude 5.00 C is at the...Ch. 15 - Two hard rubber spheres, each of mass m = 15.0 g,...Ch. 15 - Prob. 66APCh. 15 - A solid conducting sphere of radius 2.00 cm has a...Ch. 15 - Three identical point charges, each of mass m =...Ch. 15 - Each of the electrons in a particle beam has a...Ch. 15 - Protons are projected with an initial speed v0 = 9...
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