# Electric Potential - PowerPoint PPT Presentation

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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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Electric Potential

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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

• 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.

• Er = -dV/dr = -d/dr (kq/r)

• Er = kq/r2

### 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)

### Problem

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

• Find E.