Electric potential
This presentation is the property of its rightful owner.
Sponsored Links
1 / 16

Electric Potential PowerPoint PPT Presentation


  • 59 Views
  • Uploaded on
  • Presentation posted in: General

Electric Potential. Voltage. Potential of a Continuous Charge. Break the charge into small dq pieces and find the potential due to each piece, treating it as a point charge Integrate to find the potential of the whole. Example.

Download Presentation

Electric Potential

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Electric potential

Electric Potential

Voltage


Potential of a continuous charge

Potential of a Continuous Charge

  • Break the charge into small dq pieces and find the potential due to each piece, treating it as a point charge

  • Integrate to find the potential of the whole


Example

Example

  • A rod of length L has a uniform linear charge density λ. Determine the potential at a point P on the axis of the rod a distance d from one end.


Electric potential

  • dq = λ dx

  • For which dV = kdq/r

  • dV = kλdx/(d + x)

  • V =∫ol(k λdx/(d + x)

  • V = kλ∫ol[dx/(d+x)]

L

d

dx

P


Continued

Continued

  • ∫dx/(ax + b)

  • ∫du/u Let u = d +x so du = 1

  • ∫dx/(d + x) from 0 to L

  • ln(d + x) from 0 to L

  • V = kλ ln[(d+L)/d]


Problem

Problem

  • Find the electric potential at P on the central axis of the ring-shaped charge distribution of net charge Q.


Ring of charge

Ring of charge

  • Consider dq

  • r = √(x2 + R2)

  • V = k∫dq/r

  • V=k∫dq/(√(x2+R2)

  • V=k/√(x2 + R2)∫dq

  • V = kQ/√(x2 + R2)

dq

r

R

P


Problem1

Problem

  • A line of charge Q is distributed uniformly along a line of length 2a. Find the potential at a point P along the perpendicular bisector of the rod at a distance x from its center.


Picture

Picture

dq

√(x2 + y2)

P


Electric potential

  • dq = (Q/2a) dy

  • V = (kQ/2a) (∫dy/√(x2 + y2))

  • From –a to a

  • V = kQ/2a ln {√ [(a2 + y2) + a]/[√(a2 + x2) – a] }


Potential gradient

Potential Gradient

  • -dV = E dl

  • E = Exi + Eyj + Ezk

  • -dV = Exdx + Eydy + Ezdz

  • Partial Derivative

  • Ex = -dV/dx

  • Ey = -dV/dy

  • Ez = -dV/dz

  • All partial derivatives


Gradient of voltage

Gradient of Voltage

  • E = -(i dV/dx +j dV/dy + k dV/dz)

  • E = - V

  • E is the – Gradient of V


Problem2

Problem

  • From the potential at a radial distance from a point charge q V = kq/r. Find the vector electric field from this expression for V.


Answer

Answer

  • Er = -dV/dr = -d/dr (kq/r)

  • Er = kq/r2


Problem3

Problem

  • We found that for a ring of charge with a radius a and total charge Q, the potential at a point P on a ring axis a distance x from its center is

  • V = kQ/(√x2 +a2)


Problem4

Problem

  • V = (6.00 V/m) x + (4.00 V/m) y2 + (0.00 V/m) z

  • Find E.


  • Login