Chapter 17 Electric Potential

College  Physics  150   Chapter  17  –  Electric  Potential •  Electric  Potential  Energy •  Electric  Potential •  How  are  the  E-­‐‑field  and  ...
Author: Stuart Jacobs
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College  Physics  150   Chapter  17  –  Electric  Potential

•  Electric  Potential  Energy •  Electric  Potential •  How  are  the  E-­‐‑field  and  Electric  Potential  related? •  Motion  of  Point  Charges  in  an  E-­‐‑field •  Capacitors •  Dielectrics

Electric  Potential  Energy Electric  potential  energy  (Ue)  is  energy  stored  in  the  electric  field.  

• Ue  depends  only  on  the  location,  not  upon  the  path  taken  to  get  there   (conservative  force). • Ue  =  0  at  some  reference  point. • For  two  point  particles  take  Ue  =  0  at  r  =  ∞. • For  the  electric  force  

kq1q2 Ue = r

Example:  A  proton  and  an  electron,  initially  separated  by  a  distance  r,  are   brought  closer  together. (a)  How  does  the  potential  energy  of  this  system  of  charges  charge? 2 ke For  these  two  charges U = − e r

Bringing  the  charges  closer  together  decreases  r: ΔU

e

= U ef − U ei < 0

This  is  like  a  mass  falling  near  the  surface  of  the  Earth;  positive  work  is  done  by  the   field.   (b)  How  will  the  electric  potential  energy  change  if  both  particles  have   positive  (or  negative)  charges? When  q1  and  q2  have  the  same  algebraic  sign  then  ΔUe  >  0. This  means  that  work  must  be  done  by  an  external  agent  to  bring  the   charges  closer  together.

What  is  the  potential  energy  of  a  system  (arrangement)  of  point  charges?     To  calculate: Begin  by  placing  the  first  charge  at  a  place  in  space  far  from  any  other   charges.    No  work  is  required  to  do  this. Next,  bring  in  the  remaining  charges  one  at  a  time  until  the  desired   configuration  is  finished. Example:    What  is  the  potential  energy  of  three  point  charges  arranged  as  a  right   triangle?    (See  text  Example  17.2)

q2

q2

r12

q1

r12

r23

r13

q3

kq1q2 kq1q3 kq2 q3 Ue = 0 + + + r12 r13 r23

q1

r13

r23

q3

kq1q2 kq1q3 kq2 q3 + + Ue = 0 + r13 r23 r12

Are  these  the  same?

Electric  Potential

Electric  potential  is  the  electric  potential  energy  per  unit  charge.

Ue V= qtest Electric  potential  (or  just  potential)  is  a  measurable  scalar  quantity.     Its  unit  is  the  volt  (1  V  =  1  J/C).

Ue kQ = For  a  point  charge  of  charge  Q: V = qtest r When  a  charge  q  moves  through  a  potential  difference  of  ΔV,  its  potential   energy  change  is  ΔUe  =  qΔV.  

Example:  A  charge  Q  =  +1  nC  is  placed  somewhere  in  space  far  from  other   charges.    Take  ra  =  rb  =  rc  =  rd  =  1.0  m  and  re  =  rf    =  rg  =  2.0  m. f b

e

a

Q

d

c

(a)  Compare  the  potential  at  points  d  and  g. Since  Q  >  0    the  potential  at  point  d  is  greater  than   at  point  g,  it  is  closer  to  the  charge  Q.

g

(b)  Compare  the  potential  at  points  a  and  b. The  potential  at  point  a  is  the  same  as  at  point  b;  both  are  at  the  same   distance  from  the  charge  Q.

Example:  A  charge  Q  =  +1  nC  is  placed  somewhere  in  space  far  from  other   charges.    Take  ra  =  rb  =  rc  =  rd  =  1.0  m  and  re  =  rf    =  rg  =  2.0  m. f b

e

a

c

Q

d

g

(c)  Place  a  charge  of  +0.50  nC  at  point  e.    What  will  the   change  in  potential  (ΔV)  be  if  this  charge  is  moved  to   point  a?    

Ve

kQ ( 9.0 ×10 = =

Va

kQ ( 9.0 ×10 = =

9

re

ra

Nm 2 /C 2 ) (1.0 nC) 2m

9

Nm 2 /C 2 ) (1.0 nC) 1m

= +4.5 Volts = +9.0 Volts

ΔV  =  Vf  -­‐‑  Vi  =  Va  -­‐‑  Ve  =  +4.5  Volts (d)  What  is  the  change  in  potential  energy  (ΔU)  of  the  +0.50  nC  charge  ? ΔUe  =  qΔV  =  (+0.50  nC)(+4.5  Volts)=  +2.3  nJ

Example:  A  charge  Q  =  +1  nC  is  placed  somewhere  in  space  far  from  other   charges.    Take  ra  =  rb  =  rc  =  rd  =  1.0  m  and  re  =  rf    =  rg  =  2.0  m. f b

e

a

c

Q

d

(e)  How  would  the  results  of  the  previous  questions   change  if  instead  of  a  +1.0  nC  charge  there  is  a  -­‐‑1.0   nC  charge  in  its  place?

g

(a)  The potential at point d is less than the potential at point g. (b)  Unchanged (c)  -4.5 V (d)  -2.3 nJ

The  Relationship  between  E  and  V The  circles  are  called  equipotentials  (surfaces  of  equal  potential).

f

f b

e

a

Q

d

b

c e

+9  V

a

+4.5  V

g

The  electric  field  will  point  in  the  direction  of  maximum   potential  decrease  and  will  also  be  perpendicular  to  the   equipotential  surfaces.

Q

d

g

c +9  V

+4.5  V

Equipotentials   and  field  lines   for  a  dipole.

Uniform  E-­‐‑field

V1

V2

V3

V4 E

Equipotential  surfaces

ΔU e ΔV = = − Ed q

Where  d  is  the  distance  over  which   ΔV  occurs.

If  the  electric  field  inside  a  conductor  is  zero,  what  is  the  value  of  the  potential? If  E  =  0,  then  ΔV  =  0.    The  potential  is  constant! What  is  the  value  of  V  inside  the  conductor?    It  will  be  the  value  of  V  on  the   surface  of  the  conductor.

Moving  Charges When  only  electric  forces  act  on  a  charge,  its  total  mechanical  energy   will  be  conserved.  

Ei = E f Example  (text  problem  17.40):  Point  P  is  at  a  potential  of  500.0  kV  and  point  S  is  at  a   potential  of  200.0  kV.    The  space  between  these  points  is  evacuated.    When  a  charge   of  +2e  moves  from  P  to  S,  by  how  much  does  its  kinetic  energy  change?

Ei = E f K i +Ui = K f +U f

K f − K i = Ui −U f = − (U f −Ui ) = −ΔU = −qΔV = −q (Vs −Vp ) = − (+2e) ( 200.0 − 500.0 ) kV = +9.6 ×10 −14 J

Example  (text  problem  17.41):  An  electron  is  accelerated  from  rest  through  a   potential  difference.    If  the  electron  reaches  a  speed  of  7.26×106  m/s,  what  is  the   potential  difference?  

Ei = E f 0

K i +Ui = K f +U f

K f = −ΔU = −qΔV 1 2 mv f = −qΔV 2 9.11×10 kg) ( 7.26 ×10 m/s) mv ( ΔV = − =− 2q 2 (−1.60 ×10 −19 C) 2 f

= +150 Volts

−31

6

Note:  the  electron  moves   from  low  V  to  high  V.

2

Capacitors A  capacitor  is  a  device  that  stores  electric  potential  energy  by  storing  separated   positive  and  negative  charges.    Work  must  be  done  to  separate  the  charges.

+

+

+

+

+

+

+















For  a  parallel  plate  capacitor:

Parallel  plate   capacitor

E ∝Q E ∝ ΔV ∴Q ∝ ΔV

Wrijen  as  an  equality:  Q  =  CΔV,  where  the  proportionality  constant  C  is   called  the  capacitance.

Capacitors A  capacitor  is  a  device  that  stores  electric  potential  energy  by  storing  separated   positive  and  negative  charges.    Work  must  be  done  to  separate  the  charges.

+

+

+

+

+

+

+















What  is  the  capacitance  for  a  parallel   plate  capacitor?

Parallel  plate   capacitor

σ Q ΔV = Ed = d = d ε0 ε0 A ε0 A ∴Q = ΔV = CΔV d ε0 A where C = . d

Note:  C  depends  only  on  constants  and  geometrical  factors.    The  unit  of   capacitance  is  the  farad  (F).  1  F  =  1  C2/J  =  1  C/V

Example  (text  problem  17.56):  A  parallel  plate  capacitor  has  a  capacitance  of  1.20   nF.    There  is  a  charge  of  magnitude  0.800  µC  on  each  plate. (a)  What  is  the  potential  difference  between  the  plates?

Q = CΔV Q 0.800 µC ΔV = = = 667 Volts C 1.20 nF (b)  If  the  plate  separation  is  doubled,  while  the  charge  is  kept  constant,  what   will  happen  to  the  potential  difference?

Q Qd ΔV = = C ε0 A ΔV ∝ d If  d  is  doubled  so  is  the  potential  difference.

Example  (text  problem  17.100):  A  parallel  plate  capacitor  has  a  charge  of  0.020  µC   on  each  plate  with  a  potential  difference  of  240  volts.    The  parallel  plates  are   separated  by  0.40  mm  of  air. (a)  What  is  the  capacitance  of  this  capacitor?

Q 0.020 µC C= = = 8.3×10 −11 F = 83 pF ΔV 240 Volts (b)  What  is  the  area  of  a  single  plate?

ε0 A C= d Cd (83 pF ) ( 0.40 mm ) A= = ε 0 8.85 ×10 −12 C 2 /Nm 2 = 0.0038 m 2 = 38 cm 2

Dielectrics As  more  and  more  charge  is  placed  on  capacitor  plates,  there  will  come  a  point   when  the  E-­‐‑field  becomes  strong  enough  to  begin  to  break  down  the  material   (medium)  between  the  capacitor  plates. To  increase  the  capacitance,  a  dielectric  can  be  placed  between  the  capacitor   plates.

C = κ C0

ε0 A where C0 = d

and  κ  is  the  dielectric  constant.

Example  (text  problem  17.71):  A  capacitor  can  be  made  from  two  sheets  of   aluminum  foil  separated  by  a  sheet  of  waxed  paper.    If  the  sheets  of  aluminum  are   0.3  m  by  0.4  m  and  the  waxed  paper,  of  slightly  larger  dimensions,  is  of  thickness   0.030  mm  and  has  κ  =  2.5,  what  is  the  capacitance  of  this  capacitor?

ε0 A C0 = d −12 2 2 2 8.85 ×10 Nm /C 0.40 * 0.30 m ( ) ( ) = 0.030 ×10 −3m = 3.54 ×10 −8 F and C = κ C 0 = ( 2.5) (3.54 ×10 −8 F ) = 8.85 ×10 −8 F.

Energy  Stored  in  a  Capacitor A  capacitor  will   store  energy   equivalent  to  the   amount  of  work   that  it  takes  to   separate  the   charges.

The  energy  stored  in  the  electric  field  between  the   plates  is:

1 U = QΔV 2 1 2 = C ( ΔV ) 2 Q2 = 2C

}

These  are  found  by  using  Q  =   CΔV  and  the  first  relationship.

Example  (text  problem  17.79):  A  parallel  plate  capacitor  is  composed  of  two  square   plates,  10.0  cm  on  a  side,  separated  by  an  air  gap  of  0.75  mm.

(a)  What  is  the  charge  on  this  capacitor  when  the  potential  difference  is  150   volts?  

ε0 A Q = CΔV = ΔV = 1.77 ×10 −8 C d (b)  What  energy  is  stored  in  this  capacitor?  

1 U = QΔV = 1.33×10 −6 J 2