Field is the (negative of) the Gradient of Potential dU dV - - PowerPoint PPT Presentation

field is the negative of the gradient of potential
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Field is the (negative of) the Gradient of Potential dU dV - - PowerPoint PPT Presentation

Electric Potential Electric Potential energy: b F U dl elec elec a Electric Potential: b E dl V a Field is the (negative of) the Gradient of Potential dU dV


slide-1
SLIDE 1

Electric Potential

  • Electric Potential energy:
  • Electric Potential:

b elec elec a

U   F dl   

b a

V   E dl   

slide-2
SLIDE 2

Field is the (negative of) the Gradient

  • f Potential

x y z

dU F dx dU U F dy dU F dz          F

x y z

dV E dx dV V E dy dV E dz          E

slide-3
SLIDE 3

In what direction can you move relative to an electric field so that the electric potential does not change?

1)parallel to the electric field 2)perpendicular to the electric field 3)Some other direction. 4)The answer depends on the symmetry of the situation.

slide-4
SLIDE 4

Electric field of single point charge

2 ˆ

kq r E = r 

slide-5
SLIDE 5

Electric potential of single point charge

2 2

ˆ ˆ

b a b a

V kq r kq V r       

 

E dl E r r dl      

slide-6
SLIDE 6
slide-7
SLIDE 7

Electric potential of single point charge

2 2 0 by convention

ˆ ˆ .

b a b a b a b a

V kq r kq V r kq kq V V V r r kq V const r              

 

E dl E r r dl      

slide-8
SLIDE 8

Potential for Multiple Charges

1 2 3

    E E E E     

1 2 3 b a b b b a a a

V        

   

E dl E dl E dl E dl             

1 2 3

V V V V        

slide-9
SLIDE 9

Charges Q and q (Q ≠ q), separated by a distance d, produce a potential VP = 0 at point P. This means that

1) no force is acting on a test charge placed at point P. 2) Q and q must have the same sign. 3) the electric field must be zero at point P. 4) the net work in bringing Q to distance d from q is zero. 5) the net work needed to bring a charge from infinity to point P is zero.

slide-10
SLIDE 10

Finding the Electric Field due to Continuous Charge Distributions

  • When possible, use symmetry to eliminate
  • ne or more component of the electric field.
  • Define a (linear, areal or volume) density to

relate small spatial regions to small bits of charge.

  • Calculate the electric field due to each small

charge bit: .

  • Sum up (integrate) the electric fields to find

the total field.

2

r dq k dE 

slide-11
SLIDE 11

Finding the Potential due to Continuous Charge Distributions

  • When possible, use symmetry to eliminate
  • ne or more component of the electric field.
  • Define a (linear, areal or volume) density to

relate small spatial regions to small bits of charge.

  • Calculate the potential due to each small

charge bit: .

  • Sum up (integrate) the potentials to find the

total potential.

k dq dV r 

slide-12
SLIDE 12
slide-13
SLIDE 13
slide-14
SLIDE 14

Spherical shell of charge

         

2 ˆ

a constant

  • ut

in

  • ut

in

kQ r R r r R kQ V r R r kQ V R r R         E r E  