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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.

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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.
slide4
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

slide10
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.
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