Electric Potential

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# Electric Potential - PowerPoint PPT Presentation

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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## PowerPoint Slideshow about 'Electric Potential' - abel-holloway

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Presentation Transcript

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

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

dq

√(x2 + y2)

P

dq = (Q/2a) dy
• V = (kQ/2a) (∫dy/√(x2 + y2))
• From –a to a
• V = kQ/2a ln {√ [(a2 + y2) + a]/[√(a2 + x2) – a] }
• -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
• E = -(i dV/dx +j dV/dy + k dV/dz)
• E = - V
• E is the – Gradient of V
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.