1. Electric Potential Voltage

2. 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 chargeBreak 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 wholeIntegrate to find the potential of the whole

3. 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.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.

4. dq = λ dx For which dV = kdq/r dV = kλdx/(d + x) V =∫ o l (k λdx/(d + x) V = kλ∫ o l [dx/(d+x)] P d L dx

5. Continued ∫dx/(ax + b)∫dx/(ax + b) ∫du/u Let u = d +x so du = 1∫du/u Let u = d +x so du = 1 ∫dx/(d + x) from 0 to L∫dx/(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]

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

7. Ring of charge Consider dq r = √(x 2 + R 2 ) V = k∫dq/r V=k∫dq/(√(x 2 +R 2 ) V=k/√(x 2 + R 2 )∫dq V = kQ/√(x 2 + R 2 ) P R dq r

8. 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.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.

9. Picture P dq √(x 2 + y 2 )

10. dq = (Q/2a) dydq = (Q/2a) dy V = (kQ/2a) (∫dy/√(x 2 + y 2 ))V = (kQ/2a) (∫dy/√(x 2 + y 2 )) From –a to aFrom –a to a V = kQ/2a ln {√ [(a 2 + y 2 ) + a]/[√(a 2 + x 2 ) – a] }V = kQ/2a ln {√ [(a 2 + y 2 ) + a]/[√(a 2 + x 2 ) – a] }

11. Potential Gradient -dV = E dl-dV = E dl E = E x i + E y j + E z kE = E x i + E y j + E z k -dV = E x dx + E y dy + E z dz-dV = E x dx + E y dy + E z dz Partial DerivativePartial Derivative E x = -dV/dxE x = -dV/dx E y = -dV/dyE y = -dV/dy E z = -dV/dzE z = -dV/dz All partial derivativesAll partial derivatives

12. 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

13. 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.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.

14. Answer E r = -dV/dr = -d/dr (kq/r)E r = -dV/dr = -d/dr (kq/r) E r = kq/r 2E r = kq/r 2

15. 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 isWe 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/(√x 2 +a 2 )V = kQ/(√x 2 +a 2 )

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