1  

Electrostatics WS Electric Force and Field 1. Calculate the magnitude of the force between two C-60.3 µ point charges 9.3 cm

apart. 2. How many electrons make up a charge of C?0.30 µ− 3. Two charged dust particles exert a force of N102.3 2−× on each other. What will be

the force if they are moved so they are only one-eighth as far apart? 4. Compare the electric force holding the electron in orbit m)1053.0( 10−×=r around the

proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton. What is the ratio of these two forces?

5. A charge of 6.00 mC is placed at each corner of a square 0.100 m on a side.

Determine the magnitude and direction of the force on each charge. 6. A +4.75 µC and a C55.3 µ− charge are placed 18.5 cm apart. Where can a third

charge be placed so that it experiences no net force? 7. What are the magnitude and direction of the electric force on an electron in a

uniform electric field of strength CN2360 that points due east? 8. What is the magnitude of the acceleration experienced by an electron in an electric

field of ?CN750 How does the direction of the acceleration depend on the direction of the field at that point?

Electric Potential and Potential Energy 9. How much work does the electric field do in moving a C7.7 µ− charge from ground

to a point whose potential is V55+ higher?  10. An electron acquires J1045.7 16−× of kinetic energy when it is accelerated by an

electric field from plate A to plate B. What is the potential difference between the plates, and which plate is at the higher potential?

11. Two parallel plates, connected to a 200-V power supply, are separated by an air gap.

How small can the gap be if the air is not to become conducting by exceeding its breakdown value of ?mV103 6×=E

12. What is the speed of a proton whose kinetic energy is 3.2 keV?

13. A C35 µ+ point charge is placed 32 cm from an identical C35 µ+ charge. How much work would be required to move a C50.0 µ+ test charge from a point midway between them to a point 12 cm closer to either of the charges?

  2  

14. An electron starts from rest 32.5 cm from a fixed point charge with C.125.0 µ−=Q

How fast will the electron be moving when it is very far away?  15. Two point charges, C0.3 µ and C,0.2 µ− are placed 5.0 cm apart on the x axis. At

what points along the x axis is (a) the electric field zero and (b) the potential zero? Let 0=V at .∞=r

  3  

Key 1.   Use  Coulomb’s  law  to  calculate  the  magnitude  of  the  force.  

    ( ) ( )( )

269 2 21 2

22 2

3.60 10 C8.988 10 N m C 13.47N 13N

9.3 10 m

QQF k

r

−

−

×= = × ⋅ = ≈

×  

 2.   Use  the  charge  per  electron  to  find  the  number  of  electrons.  

( )6 1419

1 electron30.0 10 C 1.87 10 electrons

1.602 10 C−

−− × = ×

− ×

⎛ ⎞⎜ ⎟⎝ ⎠

 

3.   Since  the  magnitude  of  the  force  is  inversely  proportional  to  the  square  of  the  

separation  distance,  2

1F

r∝ ,  if  the  distance  is  multiplied  by  a  factor  of  1/8,  the  force  

will  be  multiplied  by  a  factor  of  64.       ( )2

064 64 3.2 10 N 2.0 NF F −= = × =  

4.   Take  the  ratio  of  the  electric  force  divided  by  the  gravitational  force.  

( )( )( )( )( )

1 2 29 2 2 192

39E 1 211 2 2 31 27

1 2G 1 22

8.988 10 N m C 1.602 10 C2.3 10

6.67 10 N m kg 9.11 10 kg 1.67 10 kg

QQkF kQQrm mF Gm mGr

−

− − −

× ⋅ ×= = = = ×

× ⋅ × ×

    The  electric  force  is  about   392.3 10× times  stronger  than  the  gravitational  force  for  the  given  scenario.    5.   Determine  the  force  on  the  upper  right  charge,  and  then  use  the  symmetry  

of  the  configuration  to  determine  the  force  on  the  other  three  charges.    The  force  at  the  upper  right  corner  of  the  square  is  the  vector  sum  of  the  forces  due  to  the  other  three  charges.    Let  the  variable   d  represent  the  0.100  m  length  of  a  side  of  the  square,  and  let  the  variable  Q  represent  the  6.00  mC  charge  at  each  corner.  

6.   Assume  that  the  negative  charge  is  d  =  18.5  cm  to  the  right  of  

the  positive  charge,  on  the  x-­‐axis.    To  experience  no  net  force,  the  third  charge  Q  must  be  closer  to  the  smaller  magnitude  charge  (the  negative  charge).    The  third  charge  cannot  be  between  the  charges,  because  it  would  experience  a  force  from  each  charge  in  the  same  direction,  and  so  the  net  force  could  not  be  zero.    And  the  third  charge  must  be  on  the  line  joining  the  other  two  charges,  so  that  the  two  forces  on  the  third  charge  are  along  the  same  line.    See  the  diagram.    Equate  the  magnitudes  of  the  two  forces  on  the  third  charge,  and  solve  for  x  >  0.  

( ) ( )

( )( )

( )

21 21 2 2 2

1 2

62

6 61 2

3.5 10 C18.5cm 116cm

4.7 10 C 3.5 10 C

QQ Q Q Qk k x d

xd x Q Q

Qx d

Q Q

−

− −

= → = → =+ −

×= = =

− × − ×

F Fr r

 

41Fr1Q

2Q 3Q

d

4Q

43Fr

42Fr

+   x  

4.7  µC   –3.5  µC  –

1Q 2Q Q

d  

  4  

7.   Use  Eq.  16–3  to  calculate  the  force.  

    ( )( )19 16 1.602 10 C 2360 N C east 3.78 10 N westqq

− −= → = = − × = ×F

E F Er

r r r  

 8.   Assuming  the  electric  force  is  the  only  force  on  the  electron,  then  Newton’s  2nd  law  may  be  used  to    

find  the  acceleration.  

   ( )( )

1914 2

net 31

1.602 10 C 750 N C 1.32 10 m s

9.11 10 kgq

m q a Em

−

−

×= = → = = = ×

×F a Er rr  

  Since  the  charge  is  negative,  the  direction  of  the  acceleration  is   opposite to the field .  

9. The work done by the electric field can be found from Eq. 17-2b.

( )( )6 4baba ba ba 7.7 10 C 55 V 4.2 10 J

WV W qV

q− −= − → = − = − − × + = ×

10. The kinetic energy gained by the electron is the work done by the electric force. Use Eq. 17-2b to calculate the potential difference.

( )

17ba

ba 19

7.45 10 J466V

1.60 10 CW

Vq

−

−

×= − = − =

− ×

The electron moves from low potential to high potential, so plate B is at the higher potential.

11. Find the distance corresponding to the maximum electric field, using Eq. 17-4b.

5 5ba ba6

200 V 6.67 10 m 7 10 m

3 10 V mV V

E dd E

− −= → = = = × ≈ ××

12. The kinetic energy of the proton is given. Use the kinetic energy to find the speed.

( )( )3 192 51

2 27

2 3.2 10 eV 1.60 10 J eV2KEKE 7.8 10 m s

1.67 10 kgmv v

m

−

−

× ×= → = = = ×

×

13. The work required is the difference in potential energy between the two locations. The test charge has potential energy due to each of the other charges, given in

Conceptual Example 17-7 as 1 2PEQQkr

= . So to find the work, calculate the

difference in potential energy between the two locations. Let Q represent the 35 Cµ charge, let q represent the 0.50 Cµ test charge, and let d represent the 32 cm distance.

[ ] [ ]initial finalPE PE

2 2 2 0.12 m 2 0.12 mkQq kQq kQq kQqd d d d

= + = +− +

14. By energy conservation, all of the initial potential energy will change to kinetic energy of the electron when the electron is far away. The other charge is fixed, and so has no kinetic energy. When the electron is far away, there is no potential energy.

( )( )

( )( ) ( )( )( )( )( )

21initial final initial final 2

9 2 2 19 7

31

7

PE KE

2 8.99 10 N m C 1.60 10 C 1.25 10 C29.11 10 kg 0.325m

3.49 10 m s

k e QE E mv

r

k e Qv

mr

− −

−

−= → = → = →

× − × − ×−= =

×

= ×

g

  5  

15. (a) Because of the inverse square nature of the electric field, any location where the field is zero must be closer to the weaker charge ( )2q . Also, in between the two charges, the fields due to the two charges are parallel to each other (both to the left) and cannot cancel. Thus the only places where the field can be zero are closer to the weaker charge, but not between them. In the diagram, this is the point labeled as “x”. Take to the right as the positive direction.

( )( )

( )

2 22 12 122

62

26 61 2

0

2.0 10 C5.0cm 22cm left of

3.0 10 C 2.0 10 C

q qE k k q d x q x

x d x

qx d q

q q

−

− −

= − = → + = →+

×= = =

− × − ×

(b)  The  potential  due  to  the  positive  charge  is  positive    

everywhere,  and  the  potential  due  to  the  negative  charge  is  negative  everywhere.    Since  the  negative  charge  is  smaller  in  magnitude  than  the  positive  charge,  any  point  where  the  potential  is  zero  must  be  closer  to  the  negative  charge.    So  consider  locations  between  the  charges  (position   1x )  and  to  the  left  of  the  negative  charge  (position   2x )  as  shown  in  the  diagram.  

   

 ( ) ( )

( )( )( )

( ) ( )( )( )

( )

6

1 2 2location 1 1 6

1 1 2 1

6

1 2 2location 2 2 6

2 2 1 2

2.0 10 C 5.0cm0 2.0cm

5.0 10 C

2.0 10 C 5.0cm0 10.0cm

1.0 10 C

kq kq q dV x

d x x q q

kq kq q dV x

d x x q q

−

−

−

−

− ×= + = → = = =

− − − ×

− ×= + = → = − = − =

+ + ×

 

So  the  two  locations  where  the  potential  is  zero  are  2.0  cm  from  the  negative  charge  towards  the  positive  charge,  and  10.0  cm  from  the  negative  charge  away  from  the  positive  charge.  

 

dx

1 0q >2 0q <

d1x2x

1 0q >2 0q <