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Electric Potential

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Electric Potential

Voltage

- 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

- 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

- ∫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]

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

- 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

- 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.

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

- 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

- 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)

- V = (6.00 V/m) x + (4.00 V/m) y2 + (0.00 V/m) z
- Find E.