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Electric
Potential
It is the degree of electrification of a
body
It determines the direction of flowof charge
between twoelectrified bodies when placed in
contact.
The charge flows froma body at higher potential to a
body having lower potential. The flowremains
continue till both potential become equal.
Physical Significance of Electric Potential:
Electric Potential:
It is an important concept in studying
electrical circuitswhere the direction of flow
of electric charges is the core part of study.
Electric Potential Energy:
It is the amount of work required tomove a
unit charge frominfinity toa point inside the
field without producing an acceleration.
Concept of Electric Potential:
It is a point used todescribe the position of
any object.
It is a "zero point" where the potential energy is zero.
Concept of Reference Point in Electric
potential “zero Potential”:
In electric Potential, the reference point is the
Earth or a point at infinity, although any point can
be used.
Reference Point:
Expression:
Statement: It is the amount of work done by
a source charge tomove a unit charge from
infinity toa point inside the field without
producing an acceleration.
Let us consider a unit test charge,
brought frominfinity toa point ‘B’
Doing that amount of work done by
source charge.
Over small displacement ‘dr’ i.e., AtoB
= − θ
Electric Potential Energy
= − 0 θ
= − 0 .
= − 0 .
= 0
⃗. = cos θ
This amount of work done store in the formof PE(U)
− = − 0 .
= − 0 .
= 0; ∞
1. Charge should be very-2 small, so it
can not disturbed electric field of
source charge
2. The motion of charge should be
uniform, so no acceleration produce
on test charge ‘ 0’
3. Charge ‘ 0’ should placed at infinity
4. Work is done by source charge ‘ ’
Important Points:
Statement:
It is the amount of work done by a
source charge tomove a unit charge
frominfinity toa point inside the field
without producing an acceleration.
=
q0
Electric Potential at any Point:
Statement:
It is the amount of work done by a
source charge tomove a unit charge
fromone point toanother point inside
the field without producing an
acceleration.
- =
q0
>
Electrostatic Potential Difference
Unit:
=
q0
volt (V) or JC-1
Define 1 volt:
If one joul work is done in moving the
positive charge of one coulomb fromone
point toanother point inside the field.
Then that potential is said tobe 1 volt
Dimensional Formula:
[M L2 T-3 A-1]
=
[M L2 T−2]
[AT]
Electrostatic Potential Difference
Let ‘ 0‘ be test charge brought frominfinity
to point ‘B’. Let ‘ ’ & ‘ ’ potential at A& B,
respectively. Then according to statement:
= θ
= 180°
= −
- =
q0
…. (1)
>
180° = -1
Electric Potential Due to a
Point Charge:
= −
4π 0
1
=
1
4π 0
= −
1
4π 0
= −
4π 0
=
+ 1
+ 1
− 1
−1
= −
4π 0
− 1
−1
=
1
=
=
4π 0
1
=
4π 0
1
−
1
∞
=
4π 0
1 ∞ = 0
= 0
ℎ ℎ
Variation of ‘V’ with ‘r’
Let ‘ 0‘ be test charge brought
frominfinity to their
respective positions one by
one i.e., 1, 2, 3,
and so on,
respectively.
= 1 + 2 + 3 + ………. (1)
Electric Potential Due to a
Group of Point Charges:
Total Potential due to all charges at point P:
Consider a systemcontaining number of
charges i.e., 1, 2, 3,
and soon.
= θ
= 180°
= −
>
180° = -1
NowPotential due to 1 :
- =
q0
….. (2)
=
1
4π 0
1
= −
1
4π 0
1
= −
1
4π 0
1
= −
1
4π 0
=
+ 1
−1= −
1
4π 0 −1
=
=
1
4π 0
1
2
1 =
=
1
4π 0
1
1
−
1
∞
=
1
4π 0
1
1
∞ = 0
= 0
= 1
Similarlly, potential due 2 :
2 = ;
= + + …….
1
=
1
4π 0
= [ + + …….. + ]
ℎ ℎ
due to a group
of charges
Prove :
q0
= ⋅
Let a charge allowed tomove fromAto B.
Then work done over small displacement ‘dl’:
= − θ … … (1)
= ⨯
= −q0 θ
W =−q0 ⋅
q0
= − ⋅ ….. (2)
Which we have toprove
Prove : Work done/unit test charge
independent of Path Followed
Since we knowthat:
q0
= − ⋅
. Δ
θ =
⇒ θ =
q0
= − ∫ θ ….. (1)
1
q0
= − ∫
=
1
4π 0
= −
= − ∫
= −
4π 0 −1
q0
= − ……. (2)
Which we have to
prove
Prove : Work done/unit test charge
over close Path is zero
Part I: Ato B, Since we knowthat:
q0
= − ⋅
. Δ
θ =
⇒ θ =
q0
= − ∫ θ ….. (1)
1
q0
= − ∫
=
1
4π 0
= −
= − ∫
= −
4π 0 −1
q0
= − ……. (2)
Part II: similarly Bto A:
q0
= − ……. (3)
Now:
q0
=
q0
+
q0
Using; (2)& (3)
q0
= 0 Which we have to
prove
Prove : Line Integral of Electric
Field over close Path is zero
Part I: Ato B, Since we knowthat:
⋅ = ∫ θ … … . (1)
. Δ
θ =
⇒ θ =
⋅ = ∫1) ⇒
= ∫
=
1
4π 0
=
= ∫
=
4π 0 −1
⋅ = − − ……. (2)
Part II: similarly Bto A:
Now:
Which we have toprove
⋅ = ⋅ + ⋅
⋅ = − − ……. (3)
⋅ = 0 Using; (2)& (3)
Expression for Potential Gradient
Since we knowthat:
( + ) - =
q0
=
q0
….. (1)
= θ
= q0 180°
= −q0
180° = − 1
= −q0
(1) implies
= −
= −
Which is required expression for
Potential Gradient
For PDF copy please write to me:
E.mail: rg68319@gmail.com
Mob: 9418076414

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Electric Potential by Dr_Raj_Paul_Guleria

  • 2. It is the degree of electrification of a body It determines the direction of flowof charge between twoelectrified bodies when placed in contact. The charge flows froma body at higher potential to a body having lower potential. The flowremains continue till both potential become equal. Physical Significance of Electric Potential: Electric Potential:
  • 3. It is an important concept in studying electrical circuitswhere the direction of flow of electric charges is the core part of study. Electric Potential Energy: It is the amount of work required tomove a unit charge frominfinity toa point inside the field without producing an acceleration. Concept of Electric Potential:
  • 4. It is a point used todescribe the position of any object. It is a "zero point" where the potential energy is zero. Concept of Reference Point in Electric potential “zero Potential”: In electric Potential, the reference point is the Earth or a point at infinity, although any point can be used. Reference Point:
  • 5. Expression: Statement: It is the amount of work done by a source charge tomove a unit charge from infinity toa point inside the field without producing an acceleration. Let us consider a unit test charge, brought frominfinity toa point ‘B’ Doing that amount of work done by source charge. Over small displacement ‘dr’ i.e., AtoB = − θ Electric Potential Energy
  • 6. = − 0 θ = − 0 . = − 0 . = 0 ⃗. = cos θ This amount of work done store in the formof PE(U) − = − 0 . = − 0 . = 0; ∞
  • 7. 1. Charge should be very-2 small, so it can not disturbed electric field of source charge 2. The motion of charge should be uniform, so no acceleration produce on test charge ‘ 0’ 3. Charge ‘ 0’ should placed at infinity 4. Work is done by source charge ‘ ’ Important Points:
  • 8. Statement: It is the amount of work done by a source charge tomove a unit charge frominfinity toa point inside the field without producing an acceleration. = q0 Electric Potential at any Point:
  • 9. Statement: It is the amount of work done by a source charge tomove a unit charge fromone point toanother point inside the field without producing an acceleration. - = q0 > Electrostatic Potential Difference
  • 10. Unit: = q0 volt (V) or JC-1 Define 1 volt: If one joul work is done in moving the positive charge of one coulomb fromone point toanother point inside the field. Then that potential is said tobe 1 volt Dimensional Formula: [M L2 T-3 A-1] = [M L2 T−2] [AT] Electrostatic Potential Difference
  • 11. Let ‘ 0‘ be test charge brought frominfinity to point ‘B’. Let ‘ ’ & ‘ ’ potential at A& B, respectively. Then according to statement: = θ = 180° = − - = q0 …. (1) > 180° = -1 Electric Potential Due to a Point Charge:
  • 12. = − 4π 0 1 = 1 4π 0 = − 1 4π 0 = − 4π 0 = + 1 + 1 − 1 −1 = − 4π 0 − 1 −1 =
  • 13. 1 = = 4π 0 1 = 4π 0 1 − 1 ∞ = 4π 0 1 ∞ = 0 = 0 ℎ ℎ
  • 14. Variation of ‘V’ with ‘r’
  • 15. Let ‘ 0‘ be test charge brought frominfinity to their respective positions one by one i.e., 1, 2, 3, and so on, respectively. = 1 + 2 + 3 + ………. (1) Electric Potential Due to a Group of Point Charges: Total Potential due to all charges at point P: Consider a systemcontaining number of charges i.e., 1, 2, 3, and soon.
  • 16. = θ = 180° = − > 180° = -1 NowPotential due to 1 : - = q0 ….. (2) = 1 4π 0 1 = − 1 4π 0 1
  • 17. = − 1 4π 0 1 = − 1 4π 0 = + 1 −1= − 1 4π 0 −1 = = 1 4π 0 1
  • 18. 2 1 = = 1 4π 0 1 1 − 1 ∞ = 1 4π 0 1 1 ∞ = 0 = 0 = 1 Similarlly, potential due 2 : 2 = ;
  • 19. = + + ……. 1 = 1 4π 0 = [ + + …….. + ] ℎ ℎ due to a group of charges
  • 20. Prove : q0 = ⋅ Let a charge allowed tomove fromAto B. Then work done over small displacement ‘dl’: = − θ … … (1) = ⨯ = −q0 θ W =−q0 ⋅ q0 = − ⋅ ….. (2) Which we have toprove
  • 21. Prove : Work done/unit test charge independent of Path Followed Since we knowthat: q0 = − ⋅ . Δ θ = ⇒ θ = q0 = − ∫ θ ….. (1)
  • 22. 1 q0 = − ∫ = 1 4π 0 = − = − ∫ = − 4π 0 −1 q0 = − ……. (2) Which we have to prove
  • 23. Prove : Work done/unit test charge over close Path is zero Part I: Ato B, Since we knowthat: q0 = − ⋅ . Δ θ = ⇒ θ = q0 = − ∫ θ ….. (1)
  • 24. 1 q0 = − ∫ = 1 4π 0 = − = − ∫ = − 4π 0 −1 q0 = − ……. (2)
  • 25. Part II: similarly Bto A: q0 = − ……. (3) Now: q0 = q0 + q0 Using; (2)& (3) q0 = 0 Which we have to prove
  • 26. Prove : Line Integral of Electric Field over close Path is zero Part I: Ato B, Since we knowthat: ⋅ = ∫ θ … … . (1) . Δ θ = ⇒ θ = ⋅ = ∫1) ⇒
  • 27. = ∫ = 1 4π 0 = = ∫ = 4π 0 −1 ⋅ = − − ……. (2)
  • 28. Part II: similarly Bto A: Now: Which we have toprove ⋅ = ⋅ + ⋅ ⋅ = − − ……. (3) ⋅ = 0 Using; (2)& (3)
  • 29. Expression for Potential Gradient Since we knowthat: ( + ) - = q0 = q0 ….. (1) = θ = q0 180° = −q0 180° = − 1
  • 30. = −q0 (1) implies = − = − Which is required expression for Potential Gradient
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