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Electrostatics

Electrostatics. Coulomb’s Law. Coulomb’s Law. F = (kq 1 q 2 )/r 2 K = 9 x 10 9 Nm 2 /C 2 Positive force means repulsion Negative force means attraction. Example problem. 4 m.

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Electrostatics

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  1. Electrostatics Coulomb’s Law

  2. Coulomb’s Law • F = (kq1q2)/r2 • K = 9 x 109 Nm2/C2 • Positive force means repulsion • Negative force means attraction

  3. Example problem 4 m • A charge q1 = 4 μC is positioned at the origin. A charge of q2 = 9 μC is positioned at x = 4 m. Where on the x-axis can a charge q3 be placed so that the force is zero? q1 q3 q2

  4. Solution • (kq1q3)/x2 = (kq2q3)/ (4-x)2 • kq1(4-x)2 = kq2 x2 • 4 μC (4-x)2 = 9 x2 • 4(16-8x+x2) = 9x2 • 5x2 +32x – 64 = 0 • x = 1.6 m

  5. Example problem • Table salt is a crystal with a simple cubic structure with Na+ and Cl- ions alternating on adjacent lattice sites. The distance between ions is a = 2.82 x 10-10 m. a)What force does Na+ experience due to one of its Cl- neighbors? B)What force does a Cl- ion experience due to a neighboring Na+? c)What force does an Na+ ion at the origin experience due to Cl- ions at (a,0,0) and (0,a,0)?

  6. Solution • A) F = kq1q2/r2 = [9 x 109 (1.6 x10-19 C)2]/(.282 x 10-10 m)2 = 2.9 x 10-9 N • B) same as a by Newton’s Third Law • C) F = F1 + F2 = 2.9 x 10-9 (i + j) N = √(2.9 x 10-9)2 + (2.9 x 10-9 )2 = 4.1 x 10-9 N

  7. Electric Field • Vector Quantity • At every point in space it has a magnitude and direction • The total electric field at any point is the sum of the electric fields due to all charges that are present • Unit: N/C • Always point away from positive charge and toward negative charge

  8. Find the force on a Ca+2 ion placed in an electric field of 800 N/C directed along the positive z-axis. Solution: F = qE q = 2e F = 2e(800N/C) 2.56 x 10-16 N Problem

  9. Electric Field • E = F/q • F = kqQ/r2 • E = kqQ/qr2 • E = kQ/r2

  10. Problem 2 • A point charge q = -8.0nC is located at the origin. Find the electric field vector at the field point x = 1.2 m and y = -1.6 m.

  11. Solution • E = - 11 N/C i and 14 N/C j

  12. Problem 3 • Four identical charges are placed on the corners of a square of side L. Determine the magnitude and direction of the electric field due to them at the midpoint of one of the square’s sides.

  13. Solution… For B and D the electric fields are the same and opposite so thy cancel each other out. B EC A P EA C D E = 2kq/(cos2θL2)

  14. Problem 4 • An electric dipole consists of +q and –q and separated by a distance of 2a. If the charges are positioned at (0,0,a and (0,0,-a) determine the electric field at a point a distance of z from the origin on the z-axis, where z >> 2a.

  15. Charge Distributions and E-fields • ΔE = k ΔQ/r2r • E = k Σi Δqi/ri2r • E = k lim Σ Δqi/ri2r = k ∫dq/r2r

  16. Linear charge distribution • Charge Q is distributed on a line of length l, the linear charge density • λ = Q/l • dq = λ dl

  17. Surface Area Charge Distribution • Charge Q is distributed on a surface area A, the surface density • δ = Q/A • dq = δ dl

  18. Volume Charge Distribution • Charge Q is distributed uniformly throughout a volume V, the volume charge density • ρ = Q/V • dq = ρ dl

  19. Problem 1 • A rod of length l has a uniform positive charge per unit length, λ, and a total charge, Q. Calculate the electric field at a point P along the axis of the rod and a distance a from one end.

  20. Problem 2 • A ring of radius a carries a uniformly distributed positive charge Q. Calculate the E field due to the ring at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring.

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