ch.24 - electric potential - an-najah staff | we challenge ... · pdf filea family of...

Dr. Iyad SAADEDDINChapter 24: Electric Potential

Electric potential Energy and Electric potential

Calculating the E-potential from E-field for different charge distributions

Calculating the E-field from E-potential

Potential of a charged isolated conductor

Conservative force Construct potential energy

Gravitational force

Spring elastic force 2

21 kxU

mghU

=

=

like

Other example experimentally discovered by physicists and engineers is Electrical force (Fe)

We also can construct a potential energy work done by the electric conservative force Wc = WFe= -∆U.

Once we have potential energy we can apply principle of conservation of mechanical energy (∆Emech = ∆K + ∆U = 0 ) for closed system involving the force and work energy theorem (W = ∆K)

Wc = -∆U24-1: What is physics

If the system changes its configuration from an initial state i to a different final state f does work W on the particles

But is a conservative force work (W) depends on the end points (i and f) and not on the path.

As for other conservative forces, the work W done by electrostatic force results in change of potential energy (∆U) of the system

When an electrostatic force acts between two or more charged particles within a system of particles, we can assign an electric potential energy U to the system.

24-2: Electric Potential Energy

or we can write

If we assume that the particle moved from infinity (taken as reference where Ui=0) to a final position within an E-field of other charged particle or particles (where Uf = U) work done by the electrostatic forces F between the particles during the move in from infinity is

UWUUUUW fi −=−=−=∆−= ∞∞ simply or Example: Near Earth’s surface the E-field has the magnitude E = 150 N/C and is directed downward. What is the change in the electricpotential energy of a released electron when the electrostatic force causes it to move vertically upward through a distance d = 520 m

JqEdU

qEddEqdFWU

1419 102.1)1)(520)(150)(106.1(180cos

cos..

−− ×−=−×−−=

°−=∆⇒−=−=−=−=∆ θ

rrrr

during the 520 m ascent, U of the electron decreases by 1.2×10-14 J

24-2: Electric Potential Energy

Work or ∆U, is dependent on the magnitude of the charge, q.We can define a quantity independent of the charge and only has an attribute of the electric field We can define the potential energy per unit charge at a point inan electric field is called the electric potential V (or simply the potential) at that point

24-3: Electric Potential

Scalar quantity

Between any two points i and f in an E-field, the potential difference ∆V is:

With ∆U = -W (∆V between two points in an E-field)

24-3: Electric PotentialSetting Ui = 0 at infinity Vi = 0 , Hence final potential Vf = V is:

(Potential at some point in an E-field)

The SI unit for potential is volt (V)

Energy is unit is

By unit conversionElectric Field units

CJV =

mJNmNJ /=⇒⋅=

( ) ( )( ) JCJCVeeV 1919 106.11106.111 −− ×=×==The Electron-Volt

mVN

mJVJ

CCNCN =⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

//

work or energy can be expressed in units of eV (unit of energy)

whereThis unit allow us to express the E-field in an other unit (V/m);

and Potential unit is VJCCJV // =⇒=

24-3: Electric Potential: work done by applied force

If a particle of charge q is moved from point i to point f in an electric field by applying a force to it, the applied force does work Wapp on the charge while the electric field (electrostatic force) does work W on it

and since

Note that ∆U or Wapp can be positive, negative, or zero depending on the signs and magnitudes of q and V.

Work done by applied force

and Work done by E-field or E-force

24-4: Equipotential Surfaces

If a particle of charge q is moved between adjacent points (i and f) that have same potential (Vi = Vf ) no work is done by the E-field or applied force to move this particle (∆V = 0 W = Wapp = 0 for any path taken; since ∆U and hence ∆V depends on end points).

Such adjacent points of same electric potential forms what is called Equipotential Surface.

Equipotential surface can be either an imaginary surface or a real, physical surface

A family of equipotential surfaces (of potentials V1, V2, V3, and V4) associated with the electric field due to some distribution of charges is show below.

100 V

80 V

60 V

40 V


For path I W = 0 (∆V = 0)

For path II W = 0 (∆V = 0)

For path III W = - q∆V = - q (V2-V1) = -q (80-100) = 20 q

For path IV W = -q∆V = - q (V2-V1) = -q (80-100) = 20 q

The work done by the E-field wile moving charged particle of charge q is:

For the paths III and IV;

If q is +ve W is +ve

and If q is –ve W is -ve

Equipotential surfaces are always perpendicular to electric field lines and thus to perpendicular to .

We can have a series of surfaces (as shown) perpendicular to and each surface has same potential as shown


For different E-field configurations, we will have different equipotential surface configurations as shown

Uniform Field Point Charge Electric Dipole

the potential between two points i and f can be calculated if we know along the path between the two points.

consider the following arbitrary path for test charge q0 in

24-5: Calculating the potential from the field

The work dW done by electrostatic force during a displacement is

integrating

but

thus

If we set Vi = 0 and Vf = VThis is the potential V at any point f relative to the zero potential at infinity.

24-5: Calculating the potential from the field: Example

two points i and f separated by a distance d in a uniform electricField. Find the potential difference Vf -Vi by moving a positive test charge q0 from i to f a) along the path if (parallel to the field direction) and b) along the path icf shown.

E-field always directed from higher potential to lower potential

Vi > Vf


a) along the path if (parallel to the field direction)

a) along the path icf

s is the length of the line cf

with sin 45° = cos 45° = d/s

As expected, same result as in (a) because the potential difference does not depend on the path

0 because ┴ (or both i and c have same potential)

Ex: Two conducting parallel plates separated by 0.3 cm, connected to a 12V battery. Calculate the E-field magnitude between the plates

mVEm

VdV

E

EdV

/104103.0

12

3

2

×=×

=∆

=⇒

−=∆⇒

−


E between two plates is uniform (constant)

Ex: Motion of a Proton in a Uniform Electric Field.

A proton released from rest, it undergo a displacement of 0.5 m in the direction of E-field as shown. Find a) ∆V and ∆U between A and B, b) the speed of the proton at point B

a)


Ex: Motion of a Proton in a Uniform Electric Field.

b) the speed of the proton at point B

b)

24-5: Calculating the potential from the field: Example – continued from previous

24-6: Potential due to a point chargeTo calculate the potential at some point P that lies

distance r from a positive charge q. consider we move a particle of charge q0 from point P (of potential V) to infinity (of potential V = 0) along radial direction as shown (simplest path)

∫∞

−=−=−r

if s.dEVVV rr0

drrqkEdrEdss.dE 2===

rr

rqk

rkq

rdrkqEdrV

rr

−=⎥⎦⎤

⎢⎣⎡−−=−=−=−⇒

∞∞

∫∫10 2

but

rqkV =⇒ potential V due to point

charge q at radial distance r

Also If q is +ve V is positive and if q is –ve V is -ve

rQkVB =

RQkVC =

rRkQE 3=

( )2232

rRR

kQEdrVVVr

RCD −=−=−=∆ ∫

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 2

2

32 R

rR

kQVD

at r > R

at r < R

For the given insulating sphere of total charge Q, Find the E-potential at points B, C, and D

Same followed for C at distance R (r = R) from center

Like for point charge

We can use the E-field to calculate VD

( ) CD VrRR

kQV +−=⇒ 2232

inside a charged solid sphere of continuous charge distribution

but

24-6: Potential due to a point charge: Example

For example: the electric potential at P fot the charges shown is

For two or more point charges, the electric potential at point P can be obtained by applying the superposition principle

V

rqqqqk

rq

rq

rq

rqkVV

ii

350919.010361099.8

)(

99

4321

4

4

3

3

2

2

1

14

1

≈×

×=

+++=

+++==⇒

−

=∑

Where ri is the distance from point P to charge qi

∑ ∑==i i

ii r

qkVV

24-7: Potential due to a group of point charge

mdrrrrr 919.02/4321 ====== 1.3 m

24-8: Potential Due to the Electric DipoleAt P located distance r from dipole center, the positive point charge (at distance r(+)) sets up potential V(+) and the negative point charge (at distance r(-)) sets up potential V(-).Then the net potential at P is:

with

or

The potential at some point P due to element charge dq is.

rdqkdV =

The total potential due to the charged object is

∫=r

dqkV

Consider a charge distributed continuously on an extended object

Remember charge densities

24-9: Potential Due to a Continuous Charge Distribution

A rod of length L located along the x axis has a total charge q and a uniform linear charge density λ. Find the electric potential at a point Plocated on the y axis a distance d from the origin

From integral tables

24-9: Potential Due to a Continuous Charge Distribution: Line of charges

For the element charge dq = λdL= λdx we have

Total potential for the rod at P is

For the uniformly charged disk of radius R and surface charge density σ. Find the electric potential at a point P located at a distance z along the perpendicular central z-axis of the disk

24-9: Potential Due to a Continuous Charge Distribution: charged disc

For the element charge dq = σdA= σ 2πR’dR’ we have

Total potential for the disc at P is From integral tables

∫−=B

A

s.dEV rrs.dEdV rr

−=⇒

dxdVEdxEdV xx −=⇒−=E

EdrdVEdrEdV rr −=⇒−=

24-9: Calculating the Field From the Potential

The potential is defined by

For E-field in x-direction

For E-field in radial-direction

xVEx ∂∂

−=zVEz ∂∂

−=

)23( 2 zyxyVEy ++−=∂∂

−=

In general, For E-field in three dimensions (Cartesian coordinates)

)(

ˆˆˆˆˆˆ

dzEdyEdxEdV

kdzjdyidxsdandkEjEiEE

zyx

zyx

++−=⇒

++=++=rr

Ex: find Ex, Ey, and Ez

xyxVEx 6−=∂∂

−=

yVEy ∂∂

−=

yzVEz −=∂∂

−=

24-9: Calculating the Field From the Potential

Ex: An electric dipole, along x-axis, centered at the origin and separated by a distance 2a. Calculate a) electric potential at point P, b) Calculate V and Ex at a point far (x >> a) from the dipole, and c) Calculate V and Ex if point P is located anywhere between the two chargesa) Assume P is located at distance x from the origin

b) 222 xaxaxfor ≈−⇒>>

and(Note that V and E are zero at x = infinity)

24-9: Calculating the Field From the Potential: Example

c) Calculate V and Ex if point P is located anywhere between the two charges

Note that at x = 0 V = 0 and

24-9: Calculating the Field From the Potential: Example – Continued from previous slide

24-9: Calculating the field From the Potential: Example

Calculate the E-field at a point P located at a distance z along the perpendicular central z-axis of a disk of radius R and has charge density σ.The electric potential at any point on the central axis of auniformly charged disk was found before

The E-potential at z is

If we have a charge q2potential at P located at r12 is

24-10 : Electric Potential Energy of a System of point charges

If we bring (from infinity by external agent) an other charge q1 and place it at point P work needed is W = q1V2

the potential energy of the system can be expressed as

The electric potential energy of a system of fixed point charges is equal to the work that must be done by an external agent to assemble the system, bringing each charge in from an infinite distance.

for system of more than two charged particles we obtain the total potential energy of the system by calculating U for every pair of charges and summing the terms algebraically

24-10 : Electric Potential Energy of a System of point charges

q1

q2

Ex: for the charges shown, a) find the total potential at P, b) if charge q3 is bring to point P (from infinity), find the change in it’s potential energy, c) the total potential energy of the system

∑ ∑==i i

iiP r

qkVV

)(2

2

1

1

rq

rqkVP +=⇒

r1

534 222 =+=r

a)

24-10 : Electric Potential Energy of a System of point charges: Example

q1

q2

Ex: for the charges shown, a) find the total potential at P, b) if charge q3 is bring to point P (from infinity), find the change in it’s potential energy, c) the total potential energy of the system

r1

534 222 =+=r

b)(Ui = 0 at ∞)

c)

24-10 : Electric Potential Energy of a System of point charges: Example – continued

For conductor in electrostatic equilibrium, if we consider two points A and B on the surface

E-field ┴ to the path connecting the two points on the surface (since E is ┴ to surface)

VA= VB

The surface of the conductor is an equipotential surface (all the surface has the same E-potential)

hence

24-11 : Potential of a Charged Isolated Conductor

If we consider two points, A on the surface

and B inside the conductor

Since Einside= 0

VA = VB = V at the surface

The E-potential is constant every whereinside the condcutor and equals to the potential on the surface

hence


For charge conducting sphere of charge Q (the figure)

a) E-potential at the surface is and is equal to potential inside the sphere

b) Outside the sphere, the E-potential is decreasing with r

c) On the other hand, the E-field inside the conductor = 0, But out side the conductor it is


Properties For conductor in electrostatic equilibrium

E-field inside the conductor is zero.

Charges always reside at the outer surface of the conductor.

The field lines are always perpendicular to surface.

Charge density is highest at smallest radius of curvature.

The surface is an equipotential surface (all surface has same E-potential).

Inside the conductor, the potential V is constant and equal to the surface value.


Ex: Two spherical conductors, in equilibrium, are connected by aconducting wire as shown in Figure. Find the ratio of the magnitudes of the electric fields at the surfaces of the spheres.

Since spheres are connected by a conductingwires they have same E-potential

But


0. =−=− ∫B

AAB dVV sE

since ∆V=0 for all paths,

Cavity within a conductor: for to points (A and B) on a conductor with cavity, if we take and path between the two points through the cavity

E-field everywhere in the cavity must be zero


SummaryElectric potential is the electric potential energy per unit charge.E-Potential is related to the electric field.All points inside a conductor are at the same potential.

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