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Physics 6BPhysics 6B

Electric Field Examples

Physics 6BPhysics 6B

Electric Field Examples

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q

q2 at the origin.b) Find the net electric force on a charge q

Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.

Find the magnitude and direction of the net electric field produced by q1 and

Find the net electric force on a charge q3=-0.6nC placed at the origin.





q2 q1

x=0 x=0.2mx=-0.3m

x






The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is awaypositive charge, and toward a negative one.



2Rkq

E = q2 q1

x=0 x=0.2mx=-0.3m


away from a

x







At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right.



2Rkq

E =


q2 q1

x=0 x=0.2mx=-0.3m

away from a

field vector that points

E1 E2

x









This is how we can put the +/- signs on the E-fields when we add them up.



2Rkq

E =


21total EEE +−= 21total

q2 q1

x=0 x=0.2mx=-0.3m

away from a


fields when

E1 E2

x









2Rkq

E =





21total EEE +−= 21total

2

9

CNm9

2

9

CNm9

total )m3.0(

)C105)(109(

)m2.0(

)C104)(109(E

2

2

2

2

−=⋅⋅

+⋅⋅

−=−−

q2 q1

x=0 x=0.2mx=-0.3m

E1 E2

x





away from a


fields when

CN

CN 500900 +−





2Rkq

E =





21total EEE +−=

(This means 400 N/C in the negative x-direction)

21total

2

9

CNm9

2

9

CNm9

total )m3.0(

)C105)(109(

)m2.0(

)C104)(109(E

2

2

2

2

−=⋅⋅

+⋅⋅

−=−−

CN

total 400E −=

q2 q1

x=0 x=0.2mx=-0.3m

Etotal

x





away from a


fields when

direction)

CN

CN 500900 +−





2Rkq

E =





21total EEE +−=

For part b) all we need to do is multiply the E-field from part a) times the new charge q

21total

2

9

CNm9

2

9

CNm9

total )m3.0(

)C105)(109(

)m2.0(

)C104)(109(E

2

2

2

2

−=⋅⋅

+⋅⋅

−=−−

CN

total 400E −= (This means 400 N/C in the negative x-direction)

q2 q3 q1

x=0 x=0.2mx=-0.3m

x





away from a


fields when

Etotal

field from part a) times the new charge q3.

CN

CN 500900 +−

direction)





2Rkq

E =





21total EEE +−=

Note that this force is to the right, which is opposite the EThis is because qup as if there are positive charges.

21total

2

9

CNm9

2

9

CNm9

total )m3.0(

)C105)(109(

)m2.0(

)C104)(109(E

2

2

2

2

−=⋅⋅

+⋅⋅

−=−−

CN

total 400E −= (This means 400 N/C in the negative x-direction)

For part b) all we need to do is multiply the E-field from part a) times the new charge q

N104.2)400)(C106.0(F 7CN9

onq3−− ⋅+=−⋅−=

q2 q3 q1

x=0 x=0.2mx=-0.3m

x





away from a


fields when

Etotal

Fon3

Note that this force is to the right, which is opposite the E-fieldThis is because q3 is a negative charge: E-fields are always set up as if there are positive charges.

CN

CN 500900 +−

direction)

field from part a) times the new charge q3.



17.28 Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F?

Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F?




The formula for electric force between 2 charges is


The formula for electric force between 2 charges is 221

elec Rqkq

F =





If both charges are doubled, we will haveelecF =



elec Rqkq

F =

221

221

Rqkq

4R

)q2)(q2(k⋅==





If both charges are doubled, we will have

So the new force is 4 times as large.

elecF =



elec Rqkq

F =

221

221

Rqkq

4R

)q2)(q2(k⋅==



17.29 Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F?

Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3







elec Dqkq

F =





We want the force to be 3 times as strong, so we can setup the force equation and solve for the new distance.




221

elec Dqkq

F =

2new

21221

Dqkq

Dqkq

3 =⋅





Canceling and cross-multiplying, we get


2newD




221

elec Dqkq

F =

2new

21221

Dqkq

Dqkq

3 =⋅

231

new D⋅=





Canceling and cross-multiplying, we get


Square-roots of both sides gives us the answer:

2newD




roots of both sides gives us the answer:

221

elec Dqkq

F =

2new

21221

Dqkq

Dqkq

3 =⋅

231

new D⋅=

DD31

new ⋅=



17.30 When two unequal point charges arthe heavier one has an acceleration a. If yof this value, how far (in terms of d) should the charges be released?

are released a distance d from one another, f you want to reduce this acceleration to 1/5

of this value, how far (in terms of d) should the charges be released?




Recall that Newton's 2nd law says that Fnet = ma.So this is really a problem about the force on the heavier charge.



So this is really a problem about the force on the heavier charge.









The formula for electric force between 2 charges is2

21elec d

qkqF =






If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong:

221

51

2new

21

old51

new

dqkq

dqkq

FF

⋅=

⋅=





elec dqkq

F =







221

51

2new

21

old51

new

dqkq

dqkq

FF

⋅=

⋅=

We cancel common terms and cross-multiply to get





elec dqkq

F =

multiply to get22

new d5d ⋅=







221

51

2new

21

old51

new

dqkq

dqkq

FF

⋅=

⋅=

We cancel common terms and cross-multiply to get

Square-root of both sides: d5dnew ⋅=





elec dqkq

F =

multiply to get22

new d5d ⋅=



17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=

is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric

axis: a) x=20 cm; b) x=1.20m; c) x= -20cm






-4nC

x=0 x=0.8m

+6nCx




This is only the magnitude. The direction is awayfrom a positive charge, and toward a negative one.


2RkQ

E =

-4nC

x=0 x=0.8m

+6nCx

one.





For part a) which direction do the E-field vectors point?




2RkQ

E =

a

field vectors point?



-4nC

x=0 x=0.8m

+6nCx

one.

-4nC

x=0 x=0.8m

+6nCx

x=0 x=0.8m



For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2




21total EEE −−=

2RkQ

E =

21total

a

x direction4nC charge #1 and the +6nC charge #2 E1

E2



one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC

x=0 x=0.8m







21total EEE −−=

2RkQ

E =

21total

2

9

CNm9

2

9

CNm9

total )m6.0(

)C106)(109(

)m2.0(

)C104)(109(E

2

2

2

2

=⋅⋅

−⋅⋅

−=−−

a


E2



one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC

x=0 x=0.8m

CN1050−







21total EEE −−=

2RkQ

E =

21total

2

9

CNm9

2

9

CNm9

total )m6.0(

)C106)(109(

)m2.0(

)C104)(109(E

2

2

2

2

=⋅⋅

−⋅⋅

−=−−

For part b) E1 points left and E2 points right

21total EEE +−=

a


E2



one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC

x=0 x=0.8m

CN1050−

b

E1

E2

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC







21total EEE −−=

2RkQ

E =

21total

2

9

CNm9

2

9

CNm9

total )m6.0(

)C106)(109(

)m2.0(

)C104)(109(E

2

2

2

2

=⋅⋅

−⋅⋅

−=−−


21total EEE +−=

2

9

CNm9

2

9

CNm9

total )m4.0(

)C106)(109(

)m2.1(

)C104)(109(E

2

2

2

2

=⋅⋅

+⋅⋅

−=−−

a


E2



one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC

x=0 x=0.8m

CN1050−

b

E1

E2

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC

CN5.312+







21total EEE −−=

2RkQ

E =

21total

2

9

CNm9

2

9

CNm9

total )m6.0(

)C106)(109(

)m2.0(

)C104)(109(E

2

2

2

2

=⋅⋅

−⋅⋅

−=−−


21total EEE +−=

2

9

CNm9

2

9

CNm9

total )m4.0(

)C106)(109(

)m2.1(

)C104)(109(E

2

2

2

2

=⋅⋅

+⋅⋅

−=−−

For part b) E1 points right and E2 points left

a


E2



one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC

x=0 x=0.8m

CN1050−

b

E1

E2

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC

CN5.312+

c

E1

E2

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC







21total EEE −−=

2RkQ

E =

21total

2

9

CNm9

2

9

CNm9

total )m6.0(

)C106)(109(

)m2.0(

)C104)(109(E

2

2

2

2

=⋅⋅

−⋅⋅

−=−−


21total EEE +−=

2

9

CNm9

2

9

CNm9

total )m4.0(

)C106)(109(

)m2.1(

)C104)(109(E

2

2

2

2

=⋅⋅

+⋅⋅

−=−−

For part b) E1 points right and E2 points left

21total EEE −+=

2

9

CNm9

2

9

CNm9

total )m0.1(

)C106)(109(

)m2.0(

)C104)(109(E

2

2

2

2

=⋅⋅

−⋅⋅

+=−−

a


E2



one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC



x=0 x=0.8m

CN1050−

b

E1

E2

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC

CN5.312+

c

E1

E2

Q1 = -4nC

x=0 x=0.8m

x

Q2 = +6nC

CN846+=

17.42 A point charge of q=+6 nC is at thepoint charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,

the point (x=0.15m,y=0m), and an identical 0.15m,0m), as shown.

Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)



Part a): TRY DRAWING THE E-FIELD VECTORS ON THE DIAGRAM


x

y

12





Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.

Etotal = 0


x

y

12

E1 E2







Etotal = 0

Part b): both vectors point away from their charge.

x

y

12



E1 E2

x

y

2 1

E1

E2





Etotal = 0

Positive x-direction



CN

2

9

CNm9

1 2400)m15.0(

)C106)(109(E

2

2

=⋅⋅

=−

CN

2

9

CNm9

2 267)m45.0(

)C106)(109(E

2

2

=⋅⋅

=−

x

y

12



E1 E2

x

y

2 1

direction



E1

E2

direction



Etotal = 0




CN

2

9

CNm9

1 2400)m15.0(

)C106)(109(E

2

2

=⋅⋅

=−

CN

2

9

CNm9

2 267)m45.0(

)C106)(109(E

2

2

=⋅⋅

=−

CN

total 26672672400E =+=

x

y

12



E1 E2

x

y

2 1

direction



E1

E2

direction

Part c): both vectors point away from their charge. We will need to use vector components to add them together.


x

yPart c): both vectors point away from their charge. We will need to use vector components to add them together.

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)



(0.15,- 0.4)





x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)




E1,y

(0.15,- 0.4)





CN

2

9

CNm9

1 5.337)m4.0(

)C106)(109(E

2

2

=⋅⋅

=−

−=

=

CN

y,1

CN

x,1

5.337E

0Ex

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)




E1,y

(0.15,- 0.4)





CN

2

9

CNm9

1 5.337)m4.0(

)C106)(109(E

2

2

=⋅⋅

=−

−=

=

CN

y,1

CN

x,1

5.337E

0Ex

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)




(0.15,- 0.4)

E2



E1,y

The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.



CN

2

9

CNm9

2 216)m5.0(

)C106)(109(E

2

2

=⋅⋅

=−

CN

2

9

CNm9

1 5.337)m4.0(

)C106)(109(E

2

2

=⋅⋅

=−

−=

=

CN

y,1

CN

x,1

5.337E

0E

triangle when you see it.

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

The 0.5m in this formula for is the distance to charge 2,

using Pythagorean theorem or 5 right


0.4m




(0.15,- 0.4)triangle when you see it.

0.3m



E2

E1,y




CN

2

9

CNm9

2 216)m5.0(

)C106)(109(E

2

2

=⋅⋅

=−

CN

2

9

CNm9

1 5.337)m4.0(

)C106)(109(E

2

2

=⋅⋅

=−

−=

=

CN

y,1

CN

x,1

5.337E

0E


x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)




0.4m




E(0.15,- 0.4)triangle when you see it.

0.3m



E1,y

E2,x

E2,y




CN

2

9

CNm9

2 216)m5.0(

)C106)(109(E

2

2

=⋅⋅

=−

CN

2

9

CNm9

1 5.337)m4.0(

)C106)(109(E

2

2

=⋅⋅

=−

−=

=

CN

y,1

CN

x,1

5.337E

0E

+=⋅+= N3N

x,2 6.129)()216(E triangle when you see it.

−=⋅+=

+=⋅+=

CN

54

CN

y,2

CN

53

CN

x,2

8.172)()216(E

6.129)()216(E

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)




0.4m





0.3m



E1,y

E2,x

E2,y




CN

2

9

CNm9

2 216)m5.0(

)C106)(109(E

2

2

=⋅⋅

=−

CN

2

9

CNm9

1 5.337)m4.0(

)C106)(109(E

2

2

=⋅⋅

=−

−=

=

CN

y,1

CN

x,1

5.337E

0E

+=⋅+= N3N


Add together the x-components and the y-components separately:

CN

CN

CN

y,total

CN

CN

CN

x,total

3.5108.1725.337E

6.1296.1290E

−=−−=

+=+=

−=⋅+=

+=⋅+=

CN

54

CN

y,2

CN

53

CN

x,2

8.172)()216(E

6.129)()216(E

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)




0.4m





0.3m



E1,y

E2,x

E2,y

components separately:




CN

2

9

CNm9

2 216)m5.0(

)C106)(109(E

2

2

=⋅⋅

=−

CN

2

9

CNm9

1 5.337)m4.0(

)C106)(109(E

2

2

=⋅⋅

=−

−=

=

CN

y,1

CN

x,1

5.337E

0E

+=⋅+= N3N


Add together the x-components and the y-components separately:

CN

CN

CN

y,total

CN

CN

CN

x,total

3.5108.1725.337E

6.1296.1290E

−=−−=

+=+=

Now find the magnitude and the angle using right triangle rules:

axisxbelow7.756.1293.510

)tan(

5.526)3.510()6.129(E CN22

total

+°=θ⇒=θ

=+=

−=⋅+=

+=⋅+=

CN

54

CN

y,2

CN

53

CN

x,2

8.172)()216(E

6.129)()216(E

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)







(0.15,- 0.4)triangle when you see it.



components separately:

Now find the magnitude and the angle using right triangle rules:

75.7º

Etotal

Part d): TRY THIS ONE ON YOUR OWN FIRST...


x

y

12

(0.15,0)(- 0.15,0)



(0,0.2)



Part d): both vectors point away from their charge. We will need to use vector components to add them together.

The 0.25m in this formula is the distance to each charge using the Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.

From symmetry, we can see that E2 will have the same components, except for +/- signs.


CN

2

9

CNm9

1 864)m25.0(

)C106)(109(E

2

2

=⋅⋅

=−

+=+=

−=−=

CN

25.020.0

CN

y,1

CN

25.015.0

CN

x,1

2.691))(864(E

4.518))(864(E

+=+= N15.0N 4.518))(864(E

Now we can add the components (the x-component should cancel out)

The final answer should be 1382.4 N/C in the positive y

+=+=

+=+=

CN

25.020.0

CN

y,2

CN

25.015.0

CN

x,2

2.691))(864(E

4.518))(864(E

CN

CN

CN

y,total

CN

CN

CN

x,total

4.13822.6912.691E

04.5184.518E

+=++=

=+−=

x

yPart d): both vectors point away from their charge. We will need to use vector components to add them together.

12

E1 E2

(0,0.2)

(0.15,0)(- 0.15,0)

The 0.25m in this formula is the distance to each charge using the Pythagorean theorem or from

5 right triangle



The final answer should be 1382.4 N/C in the positive y-direction.



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