Electric PotentialElectric Potential

VoltageVoltage

Potential of a Continuous Potential of a Continuous ChargeCharge

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

• Integrate to find the potential of Integrate to find the potential of the wholethe whole

ExampleExample

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

• dq = λ dxdq = λ dx• For which dV = For which dV =

kdq/rkdq/r• dV = kλdx/(d + x)dV = kλdx/(d + x)

• V =∫V =∫ooll(k λdx/(d + (k λdx/(d +

x)x)• V = V =

kλ∫kλ∫ooll[dx/(d+x)][dx/(d+x)]

P

d L

dx

ContinuedContinued

• ∫∫dx/(ax + b)dx/(ax + b)• ∫∫du/u Let u = d +x so du = 1du/u Let u = d +x so du = 1• ∫∫dx/(d + x) from 0 to Ldx/(d + x) from 0 to L• ln(d + x) from 0 to Lln(d + x) from 0 to L• V = kλ ln[(d+L)/d]V = kλ ln[(d+L)/d]

ProblemProblem

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

Ring of chargeRing of charge

• Consider dqConsider dq• r = √(xr = √(x22 + R + R22))• V = k∫dq/rV = k∫dq/r• V=k∫dq/(√(xV=k∫dq/(√(x22+R+R22))• V=k/√(xV=k/√(x22 + R + R22)∫dq)∫dq• V = kQ/√(xV = kQ/√(x22 + R + R22))

P

Rdq

r

ProblemProblem

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

PicturePicture

P

dq √(x2 + y2)

• dq = (Q/2a) dydq = (Q/2a) dy• V = (kQ/2a) (∫dy/√(xV = (kQ/2a) (∫dy/√(x22 + y + y22))))• From –a to aFrom –a to a• V = kQ/2a ln {√ [(aV = kQ/2a ln {√ [(a22 + y + y22) + ) +

a]/[√(aa]/[√(a22 + x + x22) – a] }) – a] }

Potential GradientPotential Gradient

• -dV = E dl-dV = E dl

• E = EE = Exx ii + E + Eyy jj + E + Ezz kk

• -dV = E-dV = Exxdx + Edx + Eyydy + Edy + Ezzdzdz

• Partial DerivativePartial Derivative

• EExx = -dV/dx = -dV/dx

• EEyy = -dV/dy = -dV/dy

• EEzz = -dV/dz = -dV/dz

• All partial derivativesAll partial derivatives

Gradient of Voltage Gradient of Voltage

• E = -(i dV/dx +j dV/dy + k dV/dz)E = -(i dV/dx +j dV/dy + k dV/dz)• E = - VE = - V• E is the – Gradient of VE is the – Gradient of V

ProblemProblem

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

AnswerAnswer

• EErr = -dV/dr = -d/dr (kq/r) = -dV/dr = -d/dr (kq/r)

• EEr r = kq/r= kq/r22

ProblemProblem

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

• V = kQ/(√xV = kQ/(√x22 +a +a22))

ProblemProblem

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

• Find E.Find E.