Electric Potential Concepts

• Focus of the week is Electric Potential
• Three concepts in electric potential:
  • Electric Potential of Electrostatic Field
  • Electric Potential Field of a Continuous Charge Body
  • Equipotential Line

Electric Potential Difference, ΔV

• Change in potential energy per unit test charge ( kecil point charge q₁) when moving from point A to B
• Equation to solve for electric potential difference (ΔV):
  • ΔU ΔV=−∫BaF⃗ ⋅ds⃗ =−∫BaE⃗ ⋅ds⃗ qt
• Units: Joules/Coulomb = Volts
• Negative sign indicates:
  • It exists from the definition, ∆U = −Wc
• Potential difference is a scalar quantity
  • There is no direction associated with it
• Tells you potentially energy difference of a charge qt between 2 points
• In Electrical Engineering, potential difference is called voltage

How Big is a Volt?

• AA Batteries: 1.5 V
• Car Batteries: 12 V
• US Outlet (AC): 120 V
• Singapore (AC): 230 V
• Distribution Power Lines: 120 V-70 kV
• High Voltage Transmission Lines: 100 kV-700 kV
• Van der Graaf: 300 kV
• Tesla Coil: 500 kV
• Lightning: 10-1000 MV

Tesla Coil

• An electrical resonant transformer circuit designed by Nikola Tesla in 1891
• Used to produce high-voltage, low-current, high frequency alternating-current electricity

E Field and Potential: Effects

• If a charged particle q is placed in an electric field E, there will be a force acting on the particle given by F = qẺ
• If q is moved over a displacement of electric potential ΔV, then the potential change is ΔU = qΔV
• If there is not external acting on the particle over the displacement, kinetic energy change ΔK = −ΔU
• Wext refers to work done by the external force
  • If there is an external force acting to the system, we have ΔK + ΔU = Wext
  • Total energy of the system is no longer conserved, but is increased due to Wext acting onto the system
  • If the kinetic energy is to remain constant over the displacement (ΔΚ = 0), the Wext = ∆U = q∆V
  • Work done by the external force converts to the potential energy stored in the system

Concept Question 1.1: Motion of Charged Objects

• Two oppositely charged are released from rest in an electric field.
• Answer:
  • The positively charged object will move from higher to lower electric potential
  • The negatively charged object will move from lower to higher electric potential
• ΔV < 0 ⇒ ΔU = qΔV < 0 (For the positively charged object)
• ΔV > 0 ⇒ ΔU = qΔV < 0 (For the negatively charged object)

Potential Created by Point Charge

• Equation to solve potential of point charge:
  • E = kQr^/r^2
  • ds = drr^ + rddθθ^
• Formula:
  • ΔV = VB - VA = -∫BA ⃗E ⋅ ds⃗
  • =-∫BAkQq^/r^2⋅(drr^+rddθθ^)=-kQ∫BAdr/r^2
  • =kQ(1/rB - 1/rA )
• Formula to solve VQ(r) when reference is taken at rA = ∞:
  • VQ(r) = kQ/r

Potential by Multiple Point-Charge

• The net electric field at point P is the superposition of all the electric fields contributed by individual point charges
• Thus, electric potential at P due to N discrete point charges can be solved using formula:
  • V(⃗r )-V(∞)=keΣNs=1qs/ |⃗r p-⃗r s|
• where V(∞) ≡ 0

Concept Question 1.2: Two Point Charges Equation

• The work done is moving a positively charged object that starts from rest at infinity and ends at rest at the point P midway between two charges of magnitude +Q and -Q
• Answer:
  • Work from ∞ to P is zero (C)
• The potential at ∞ is zero
• The potential at P is zero because equal and opposite potentials are superimposed from the two point charges
  • V is a scalar, not a vector

Electric Potential Field of a Continuous Charge Body

• Method 1: From continuous charge distribution
• Method 2: Using Gauss's Law to Find Electric Potential from Electric Field

Method 1: Continuous Charge Distributions

• Break the continuous charge distribution into infinitesimal charged elements, each of charge dq
• Formula to solve electric potential difference between infinity and P due to dq:
  • dV = Vdq(P) – Vdq(∞) = ke (dq/r)
    • Note: r is the distance from dq to the point of interest, P
• Formula for solving the superposition principle:
  • V(P) – V(∞) = ke ∫source (dq/r)
    • The infinity is usually set as the reference point to have zero potential: V(∞) = 0
• Thus, the potential at point P due the continuous charge body:
• V(P) = ke ∫source dq/r
• Steps to Calculating Electric Potential Difference for Continuous Distributions:
  • Choose V(∞) = 0
  • From the continuous charge, choose a small dq
    • Identify the dq position from the origin, rs
  • Identify and relate dq with geometry
    • dq = {λds →1D; σdA →2D; ρdV →3D}
  • Identify the position of point of interest P, rp
  • Calculate dq to point P distance, | p- s |
  • Electric potential due to dq is dV = ke (dq/ | r p- rs|)
  • Identify the limits of integral and integrate:
  • V(r) = ke ∫source (dq/| r p- rs|)

Worked Example: Uniform Charged Ring

• Consider a uniformly charged ring total charge Q.
• Find the electric potential difference between infinity and a point P along the symmetric axis a distance z from the center of the ring.
  • λ =Q/2πR dq = λdl = λRdφ
  • Choose V(∞) = 0
  • r' = Rr^
  • r = zk^
  • |r - r'| = √R^2 + z^2
  • dV = ke (dq / |r - r|)=1/4πε0(dq/√R^2 + z^2 )=1/4πε0( λR dφ/√R^2 + z^2 ) = λR/4πε0√R^2 + z^2∫2π0dφ=keλR(2π)/√R2 + z2 = keQ/√R2 + z2

Case Problem 2.1: Thin Rod Solution

• A thin rod extends along the x-axis from x = -l/2 to x = l/2.
• The rod carries a uniformly distributed positive charge +Q
• Calculate the electric potential difference between infinity and at a point P along the x-axis.
  • ∫dx'/(x-x') = - ln|x - x']
  • V(∞) = 0
  • dV =( dq)/(4πε0r) = (1)/(4πε0) (λdx')/(x-x')
  • V =λ/4πε0∫−l/2l/2dx'/(x−x')= -λ/4πε0 [ln|x - x' ⁄ ]l/2−l/2
  • V = (-λ/4πε0)*ln[(x - L/2)/(x + L/2)] > 0

Case problem 2.1: Extreme case

• (x >> l), the electrical potential of P should look like that of a point charge.
  • =λ/4πε0(− l/x)=(Q/4πε0x
  • ln[1− l/(2x)]≈− l/(2x)+…, in the limit when x ≫ l
  • ln[1+ l/(2x)]≈ l/(2x) +…
  • ln[x−l/2/(x+l/2)] = ln[1− l/(2x)/(1+ l/(2x))]= ln[1− l/(2x)] − ln[1+ l/(2x)] ≈ −l/(2x)− l/(2x) =−l/x

Summary: Electric Field and Potential

• A point charge q creates an electric field and around it

• ⃗E(r^)=keq/r^2r^;V(r)=keq/r

• Use superposition for systems of charges

• Set of discrete charges:

• -⃗E(r^)=ke∑qi/|⃗r p-⃗r s|3(⃗r p-⃗r s)

• V( ⃗r )-V(∞)=keΣqi/|⃗r p-⃗r s|

• Continuous charges:

• -⃗E(r^)=ke∫dq/|⃗r p-⃗r s|3(⃗r p-⃗r s)

• V( ⃗r )-V(∞)=ke∫dq/|⃗r p-⃗r s|

Method 2: Using Gauss's Law

• It you already know electric field (usually Gauss's Law)
• Can compute electric potential difference using V( ⃗r )-V(∞)= -∫BAE⃗ ⋅ ds⃗

Method 2: Find Electric Potential Difference from Electric Field (Gauss's Law)

• The relation between electric field and potential can be used to calcuate electric potential Vusing the following relation: VB - VA = -∫BAE⃗ ⋅ ds⃗
• Find E field using Gauss's Law if the charge distribution is highly symmetric, and integrate to find potential difference between A and B
• This method is useful for highly symmetrical charge distribution!

Worked Example: Nested Spherical Shells

• Two spherical shells have equal but opposite charge
• Set V(∞) = 0, helps recognize that potential, and potential of point outside can be found inside in a step-wise process
• V(r) – V(∞) for the regions
  • r>b
  • a a, E = 0 (E field inside a conductor is always zero) will result in:
  • V(r) – V(a) = -∫ar dr = 0 where V(r) = V(a) = Q/4πε0(a - 1/b)
    • Potential is constant because E = 0, but the potential is NOT ZERO for

Worked example: Coaxial cylinders

• A very long thin uniformly charged cylindrical shell (length h and radius a) carrying a positive charge +Q is surrounded by a thin uniformly charged cylindrical shell (length h and radius b ) with negative charge -Q, as shown in above. May ignore edge effects. Find (b) -V(a).
• Very long thin uniformly charged cylindrical shell is highly symmetric. Can be calculated using Gauss's Law to find the E field in between the 2 shells.
• Note the region a and r > b have zero E field
  • ∫∫E dA = ∫2πrh'E = qenc/ε0 = σ2πdh'/ε0
  • E⃗ = σε0r^; σ = Q/2παh
  • • = -Q at the outer shell does not contribute E (a
  • ΔV = V(b) – V(a) = −∫ba ⃗E ⋅ ds⃗ = -∫baσα/ε0rdr =- σα/ε0ln a/b=-Q/2πε oh lna/b
• Result shows the electric field points in the direction of decreasing electric potential

Case Problem 2.2: Charge Slab

• Find V(x₁) - V(0); xa > d

• First step: find electric field both inside and outside using symmetry

• Next: The symmetry is planar so we use Gaussian surface (cylinders of cross-section A and height 2x and place symmetrically w.r.t. x = 0)

• By symmetry magnitude of the field will be the same on the left as on the right of slab but will point in the opposite direction. There are two distinct regions of space, depending on space inside or outside of the slab

• Solution:

  • E⃗(x)=px^/ε0 for _d - { (pd/ε0)i^ for x≥d/

• V(xa) - V(o)=- SxA0E(x)dx= -s0dpx/ε0dx- sd xa pd/ε0dx = - ε 02 pd2 -pd/ε0(xa-d) - ε0(xa-d/2) for xs d

Generalization: Concept 3: Equipotential Line

• All points on equipotential curve are at same potential

• Each curve represented V(x,y) by = constant

• E is perpendicular to all equipotential lines

Properties of Equipotential Line

• E field lines point from high to low electric potential
• E field lines is always perpendicular to equipotential lines
• The E field has no component along equipotential lines
• The electrostatic force does zero work to move a charged particle along an equipotential line

Conductors are Equipotential Surfaces

• Charged conductors are equipotential objects at electrostatic equilibriums or no charges moving on surface!
• Formula:
  • V1 = keq/r

Application: Lightning Rod

• Lightning rods have a sharp point at the top
• A small radius curve of a conductor can setup a higher local electric field
• A very pointed conductor has a large charge concentration at the point
• The conductor has a force large enough to transfer change on/off conductor
• The large negative charges created are storm coulds, induce opposite charge in a building and could result in lightning
• The induced (positive) is discharged away continually reducing the change of more dramatic lightning strikes
• BUT, if it does strikes the charge will still flow into the earth