MSTC Physics CMSTC Physics CStudy GuideStudy Guide

Chapter 19Chapter 19

Sections 9 & 10Sections 9 & 10

Net FluxNet Flux

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

r

Net FluxNet Flux

Coulomb’s law Coulomb’s law says the says the magnitude of the magnitude of the electric field is kq/relectric field is kq/r22 anywhere on the anywhere on the surfacesurface

q

r

Net FluxNet Flux

E field lines point E field lines point radially outward radially outward and are and are perpendicular to perpendicular to the spherical the spherical surface at each surface at each pointpoint

q

rE

Net FluxNet Flux

For each element For each element of area, of area, ΔΔA, on the A, on the surface of the surface of the sphere sphere ΦΦ = E = EΔΔAA

For the entire For the entire surface surface ΦΦ = ∫EdA = = ∫EdA = E∫dA = kq/rE∫dA = kq/r22 (4 (4ππrr22) ) = 4= 4ππkqkq

q

r

Net FluxNet Flux

If k = 1/4If k = 1/4πεπε where where εε = permittivity of = permittivity of free space = 8.85 x free space = 8.85 x 1010-12-12 C C22/Nm/Nm22 then then ΦΦ = q/ = q/εε

Note: result is Note: result is independent of rindependent of r

q

r

Net FluxNet Flux

The electric flux is The electric flux is proportional to the proportional to the charge inside a charge inside a closed (Gaussian) closed (Gaussian) surfacesurface

ΦΦ αα q qinin

q

r

Net FluxNet Flux

Consider several Consider several closed surfaces closed surfaces surrounding a surrounding a charge +qcharge +q

S1S1

S2S2+q+q

Net FluxNet Flux

ФФ through S2 = q/ through S2 = q/єє

Know Know ФФ αα # E field # E field lines that pass lines that pass through the surfacethrough the surface

So So ФФ through S1 = through S1 = q/q/єє also also

S1S1

S2S2+q+q

Net FluxNet Flux

Net flux through Net flux through any closed surface, any closed surface, regardless of regardless of shape, is given by shape, is given by qqin/in/єє where q where qinin is the is the

charge enclosedcharge enclosed

S1S1

S2S2+q+q

Net FluxNet Flux

+q+q

Consider a point Consider a point charge outside a charge outside a closed surfaceclosed surface

# of E field lines # of E field lines that enter equals that enter equals the # of E field the # of E field lines that exitlines that exit

Net FluxNet Flux

+q+q

Net Net ФФ through a through a closed surface is closed surface is zerozero

Net flux through Net flux through any closed surface, any closed surface, regardless of regardless of shape, that shape, that contains no charge contains no charge is zerois zero

Net FluxNet Flux

Consider a surface that encloses several Consider a surface that encloses several point chargespoint charges

The E field due to many charges is the The E field due to many charges is the vector sum of the E fields produced by the vector sum of the E fields produced by the individual chargesindividual charges

∫ ∫ E ● dA = ∫ (E1 + E2 + …) ● dAE ● dA = ∫ (E1 + E2 + …) ● dA

ConsiderConsider

S1S1

S2S2

S3S3

ФФ through S1 due to through S1 due to q2 and q3 is zero q2 and q3 is zero since each E field line since each E field line that enters, exitsthat enters, exits

Ф through S1 = q1/Ф through S1 = q1/єє

ФФ through through

S2 = (q2+q3)/S2 = (q2+q3)/єє

q1

q2

q3

ConsiderConsider

S1S1

S2S2

S3S3

ФФ through S3 is zero through S3 is zero since it contains no since it contains no chargechargeq1

q2

q3

Gauss’s LawGauss’s Law Net electric flux through any closed Gaussian surface Net electric flux through any closed Gaussian surface

is equal to the net charge inside the surface divided by is equal to the net charge inside the surface divided by єє

ФФ = ∫ E ● dA = q = ∫ E ● dA = qinin / / єє

Note: qNote: qinin = charge inside = charge inside E = electric field includingE = electric field including contributions from inside andcontributions from inside and outsideoutside

Sample ProblemSample Problem

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

S1S1

S2S2

S3S3

S4S4

-

-Q

-2Q

+Q

Sample ProblemSample Problem

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

Sample ProblemSample Problem

Five charges are placed in a closed box. Five charges are placed in a closed box. Each charge (except the first) has a Each charge (except the first) has a magnitude which is twice that of the magnitude which is twice that of the previous one placed in the box. If all previous one placed in the box. If all charges have the same sign and if (after charges have the same sign and if (after all the charges have been placed in the all the charges have been placed in the box) the net electric flux through the box is box) the net electric flux through the box is 4.8 x 104.8 x 1077 Nm Nm22/C, what is the magnitude of /C, what is the magnitude of the smallest charge in the box? Does the the smallest charge in the box? Does the answer depend on the size of the box?answer depend on the size of the box?

Gauss’s LawGauss’s Law

We use Gauss’s Law to calculate E for a We use Gauss’s Law to calculate E for a given charge distributiongiven charge distribution

Gauss’s law works best if there is Gauss’s law works best if there is symmetry in the charge distributionsymmetry in the charge distribution

Choose a Gaussian surface so it has the Choose a Gaussian surface so it has the same symmetry as the charge distributionsame symmetry as the charge distribution

Sample ProblemSample Problem

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

Sample ProblemSample Problem

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

Sample ProblemSample Problem

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

Sample ProblemSample Problem

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

Sample ProblemSample Problem

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

Sample ProblemSample Problem

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

Sample ProblemSample Problem

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

Sample ProblemSample Problem

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