Electric Potential

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 charge • Integrate 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.

4. 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

5. 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]

6. Problem • 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 = √(x2 + R2) • V = k∫dq/r • V=k∫dq/(√(x2+R2) • V=k/√(x2 + R2)∫dq • V = kQ/√(x2 + R2) dq r R P

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.

9. Picture dq √(x2 + y2) P

10. dq = (Q/2a) dy • V = (kQ/2a) (∫dy/√(x2 + y2)) • From –a to a • V = kQ/2a ln {√ [(a2 + y2) + a]/[√(a2 + x2) – a] }

11. Potential Gradient • -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

12. Gradient of Voltage • E = -(i dV/dx +j dV/dy + k dV/dz) • E = - V • E 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.

14. Answer • Er = -dV/dr = -d/dr (kq/r) • Er = kq/r2

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 is • V = kQ/(√x2 +a2)

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