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MSTC Physics C Study Guide Chapter 19 Sections 9 & 10.

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Presentation on theme: "MSTC Physics C Study Guide Chapter 19 Sections 9 & 10."— Presentation transcript:

1 MSTC Physics C Study Guide Chapter 19 Sections 9 & 10

2 Net Flux  Consider a positive point charge q located at the center of a sphere of radius r q r

3 Net Flux  Coulomb’s law says the magnitude of the electric field is kq/r 2 anywhere on the surface q r

4 Net Flux  E field lines point radially outward and are perpendicular to the spherical surface at each point q r E

5 Net Flux  For each element of area, ΔA, on the surface of the sphere Φ = EΔA  For the entire surface Φ = ∫EdA = E∫dA = kq/r 2 (4πr 2 ) = 4πkq q r

6 Net Flux  If k = 1/4πε where ε = permittivity of free space = 8.85 x C 2 /Nm 2 then Φ = q/ε  Note: result is independent of r q r

7 Net Flux  The electric flux is proportional to the charge inside a closed (Gaussian) surface Φ α q in q r

8 Net Flux  Consider several closed surfaces surrounding a charge +q S1 S1 S2 S2 +q

9 Net Flux  Ф through S2 = q/є  Know Ф α # E field lines that pass through the surface  So Ф through S1 = q/є also S1 S1 S2 S2 +q

10 Net Flux  Net flux through any closed surface, regardless of shape, is given by q in/ є where q in is the charge enclosed S1 S1 S2 S2 +q

11 Net Flux +q  Consider a point charge outside a closed surface  # of E field lines that enter equals the # of E field lines that exit

12 Net Flux +q  Net Ф through a closed surface is zero  Net flux through any closed surface, regardless of shape, that contains no charge is zero

13 Net Flux  Consider a surface that encloses several point charges  The E field due to many charges is the vector sum of the E fields produced by the individual charges ∫ E ● dA = ∫ (E1 + E2 + …) ● dA ∫ E ● dA = ∫ (E1 + E2 + …) ● dA

14 Consider S1 S1 S2 S2 S3 S3  Ф through S1 due to q2 and q3 is zero since each E field line that enters, exits  Ф through S1 = q1/є  Ф through S2 = (q2+q3)/є S2 = (q2+q3)/є q1 q2 q3

15 Consider S1 S1 S2 S2 S3 S3  Ф through S3 is zero since it contains no charge q1 q2 q3

16 Gauss’s Law  Net electric flux through any closed Gaussian surface is equal to the net charge inside the surface divided by є Ф = ∫ E ● dA = q in / є  Note: q in = charge inside E = electric field including E = electric field including contributions from inside and contributions from inside and outside outside

17 Sample Problem Four closed surfaces, S1 through S4, together with the charges -2Q, +Q, and –Q are sketched in the figure. Find the electric flux through each surface. S1 S1 S2 S2 S3 S3 S4 S4 - -Q -2Q +Q

18 Sample Problem  A point charge of 12 μC is placed at the center of a spherical shell of radius 22 cm. What is the total electric flux through a) the entire surface of the shell and b) any hemispherical surface of the shell? c) Do the results depend on the radius?

19 Sample Problem Five charges are placed in a closed box. Each charge (except the first) has a magnitude which is twice that of the previous one placed in the box. If all charges have the same sign and if (after all the charges have been placed in the box) the net electric flux through the box is 4.8 x 10 7 Nm 2 /C, what is the magnitude of the smallest charge in the box? Does the answer depend on the size of the box?

20 Gauss’s Law  We use Gauss’s Law to calculate E for a given charge distribution  Gauss’s law works best if there is symmetry in the charge distribution  Choose a Gaussian surface so it has the same symmetry as the charge distribution

21 Sample Problem Find E due to an isolated point charge q.

22 Sample Problem  Find E outside and inside a thin shell of radius r with total charge Q.

23 Sample Problem Find E outside and inside a solid insulating sphere of radius r with total charge Q.

24 Sample Problem A long straight wire has a uniform charge per unit length λ. Find E at a point near the wire.

25 Sample Problem Consider a thin spherical shell of radius 14 cm with a total charge of 32 μC distributed uniformly on its surface. Find the electric field for the following distances from the center of the charge distribution: a) r = 10 cm and b) r = 20 cm.

26 Sample Problem An insulating sphere is 8 cm in diameter, and carries a +5.7 μC charge uniformly distributed throughout its interior volume. Calculate the charge enclosed by concentric spherical surfaces with the following radii: a) r = 2 cm and b) r = 6 cm.

27 Sample Problem  A solid sphere of radius 40 cm has a total positive charge of 26 μC uniformly distributed throughout its volume. Calculate the electric field intensity at the following distances from the center of the sphere: a) 0 cm, b) 10 cm, c) 40 cm, d) 60 cm.

28 Sample Problem A uniformly charged, straight filament 7 m in length has a total positive charge of 2 μC. An uncharged cardboard cylinder 2 cm in length and 10 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find a) the electric field at the surface of the cylinder and b) the total electric flux through the cylinder.


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