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

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

Net Flux Coulomb’s law says the magnitude of the electric field is kq/r2 anywhere on the surface r q Recall that F = kqq/r2 and E = F/q so E = (kqq/r2)/q or kq/r2

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

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/r2 (4πr2) = 4πkq r q

Net Flux If k = 1/4πε where ε = permittivity of free space = 8.85 x 10-12 C2/Nm2 then Φ = q/ε Note: result is independent of r r q

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

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

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 S2 +q +q

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

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

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

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

Consider S1 S2 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 q1 q2 q3

Consider Ф through S3 is zero since it contains no charge S1 S2 S3 q1

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 = qin / є Note: qin = charge inside E = electric field including contributions from inside and outside

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 S2 S3 S4 - -2Q -Q +Q S1 Φ= (-2Q + Q)/є = -Q/є S2 Φ=(+Q-Q)/є = 0 S3 Φ=(-2Q+Q-Q)/є = -2Q/є S4 Φ=0

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? Φ=12µ/є = 1.36 x 106 Φ = ½ Φ through sphere since Φ is proportional to the number of E field lines that pass through and only half as many do No – same number of E field lines pass through

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 107 Nm2/C, what is the magnitude of the smallest charge in the box? Does the answer depend on the size of the box? Φ= q/є 4.8 x 107 = (Q+2Q+4Q+8Q+16Q)/є Q=1.37 x 10-5 b) no

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

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

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

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

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

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.

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.

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.

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.