Gauss' Law For Electricity
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Gauss' Law for Electricity • The electric flux out of any closed surface is proportional to the total charge enclosed within the surface. • The integral form of Gauss’ Law finds application in calculating electric fields around charged objects. • In applying Gauss' law to the electric field of a point charge, one can show that it is consistent with Coulomb’s law. • While the area integral of the electric field gives a measure of the net charge enclosed,
The divergence of the electric field gives a measure of the density of sources.
.
It also has implications for the conservation of charge
QUICK QUIZ 1
Consider the electric fields E1 at a distance r1 and E2 at a distance r2 from the center of a uniformly charged insulating sphere of total positive charge Q and radius a. If the radius of the sphere were suddenly doubled and the total charge Q were now uniformly spread throughout this new volume, the electric fields at the same locations, r1 and r2, would be a) E1 and E2, b) 2E1 and E2, c) E1/2 and E2, d) E1/8 and E2, e) E1/2 and 2E2, or f) E1/8 and 8E2.
Answer (d) The electric field inside the sphere is proportional to the charge density and the distance from the center of the sphere. The distance does not change but the charge density is reduced by a factor of 8 (since the volume increases by a factor of 8) when the radius increases by a factor of 2. Outside the sphere, since the total charge has not changed and the distance has not changed, Gauss’s law shows that the electric field must remain the same.
QUICK QUIZ 2 A conducting spherical shell (below) is concentric with a solid conducting sphere. Initially, each conductor carries zero net charge. A charge of +2Q is placed on the inner surface of the spherical shell. After equilibrium is achieved, the charges on the surface of the solid sphere, q1, the inner surface of the spherical shell, q2, and the outer surface of the spherical shell, q3, are a) q1 = Q, q2 = +Q, q3 = +Q b) q1 = 0, q2 = 0, q3 = +2Q c) q1 = 0, q2 = +2Q, q3 = 0 d) q1 = +Q, q2 = Q, q3 = +3Q
Answer (b) •
(b). Since the electric field inside the spherical conducting shell is zero, the net charge contained in a gaussian surface inside the shell that encloses the charges q1 and q2 will be zero. Since the solid sphere had initially zero net charge, from charge conservation, it will remain with zero net charge. Therefore, q1 must be zero, and since q1 + q2 = 0, q2 must also equal zero. The +2Q excess charge on the spherical shell will migrate to the outer surface of the shell. This is consistent with the fact that like charges repel and will spread as far apart from one another as possible.
Quick Quiz 3 Suppose the radius of a sphere (radius 1.00 m, with a charge of +1.00 μC at its center) is changed to 0.500 m. What happens to the flux through the sphere and the magnitude of the electric field at the surface of the sphere? (a) The flux and field both increase. (b) The flux and field both decrease. (c) The flux increases and the field decreases. (d) The flux decreases and the field increases. (e) The flux remains the same and the field increases. (f) The flux decreases and the field remains the same.
Answer: (e). The same number of field lines pass through a sphere of any size. Because points on the surface of the sphere are closer to the charge, the field is stronger.
Quick Quiz 4. In a charge-free region of space, a closed container is placed in an electric field. A requirement for the total electric flux through the surface of the container to be zero is that (a) the field must be uniform (b) the container must be symmetric (c) the container must be oriented in a certain way (d) The requirement does not exist – the total electric flux is zero no matter what.
Answer: (d). All field lines that enter the container also leave the container so that the total flux is zero, regardless of the nature of the field or the container.
Quick Quiz 5 If the net flux through a gaussian surface is zero, the following four statements could be true. Which of the statements must be true? (a) There are no charges inside the surface. (b) The net charge inside the surface is zero. (c) The electric field is zero everywhere on the surface. (d) The number of electric field lines entering the surface equals the number leaving the surface.
Answer: (b) and (d). Statement (a) is not necessarily true because an equal number of positive and negative charges could be present inside the surface. Statement (c) is not necessarily true, a nonzero electric field exists everywhere on the surface, but the charge is not enclosed within the surface; thus, the net flux is zero.
Quick Quiz 6 Consider the charge distribution shown in the figure. The charges contributing to the total electric flux through surface S’ are (a) q1 only (b) q4 only (c) q2 and q3 (d) all four charges (e) none of the charges
Answer: (c). The charges q1 and q4 are outside the surface and contribute zero net flux through S’.
Quick Quiz 7 Again consider the charge distribution shown in this figure. The charges contributing to the total electric field at a chosen point on the surface S’ are (a) q1 only (b) q4 only (c) q2 and q3 (d) all four charges (e) none of the charges
Answer: (d). We don't need the surfaces to realize that any given point in space will experience an electric field due to all local source charges
Problem1 • A spherical shell is placed in a uniform electric field. Find the total electric flux through the shell. • Answer: zero
Problem 2 • (a) A point charge q is located a distance d from an infinite plane. Determine the electric flux through the plane due to the point charge. (b) A point charge q is located a very small distance from the center of a very large square on the line perpendicular to the square and going through its center. Determine the approximate electric flux through the square due to the point charge. (c) Explain why the answers to parts (a) and (b) are identical. • Answer: (a) q/2ε0; (b) q/2ε0; (c) The plane and the square look the same to the charge.
Gauss' Law for Magnetism The net magnetic flux out of any closed surface is zero. This amounts to a statement about the sources of magnetic field. For a magnetic dipole, any closed surface the magnetic flux directed inward toward the south pole will equal the flux outward from the north pole. The net flux will always be zero for dipole sources. If there were a magnetic monopole source, this would give a non-zero area integral. The divergence of a vector field is proportional to the point source density, so the form of Gauss' law for magnetic fields is then a statement that there are no magnetic monopoles.
Problem 3 The following charges are located inside a submarine: 5.00μC, -9.00 μC, 27.0 μC, and -84.0 μC. (a) Calculate the net electric flux through the hull of the submarine. (b) Is the number of electric field lines leaving the submarine greater than, equal to, or less than the number entering it?
Problem 4. The electric field everywhere on the surface of a thin spherical shell of radius 0.750m is measured to be 890N/C and points radially toward the center of the sphere. (a) What is the net charge within the sphere’s surfaces? (b) What can you conclude about the nature and distribution of the charge inside the spherical shell?
Faraday's Law of Induction The line integral of the electric field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop. This line integral is equal to the generated voltage or emf in the loop, so Faraday's law is the basis for electric generators. It also forms the basis for inductors and transformers.
QUIZ For a long solenoid with closely spaced turns, each coil will a) exert an attractive force on the next adjacent coil, b) exert a repulsive force on the next adjacent coil, c) exert zero force on the next adjacent coil, or d) exert an attractive or repulsive force on the next adjacent coil, depending on the current direction in the solenoid.
(a). Because the coils are closely spaced, the distance between adjacent coils will be much smaller than the radius, r, of each coil and the forces between the coils can be approximated as the same as the force between two parallel wires of length 2pr. Since the currents are in the same direction, the force will be attractive.
Ampere's Law In the case of static electric field, the line integral of the magnetic field around a closed loop is proportional to the electric current flowing through the loop. This is useful for the calculation of magnetic field for simple geometries.
Q. A wire carrying a current of 5.00 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at the center of the loop is 10.0 μT, what is the required radius? Answer: 31.4 cm
Q. A cylindrical conductor of radius R = 2.50 cm carries • a current of I = 2.50 A along its length; this current is • uniformly distributed throughout the cross-section • of the conductor. • (a) Calculate the magnetic field midway along the radius of the wire (that is, at r = R/2). (b) Find the distance beyond the surface of the conductor at which the magnitude of the magnetic field has the same value as the magnitude of the field at r = R/2. Answer: (a) 10.0 μT; (b) 2.50 cm
Q. A current path shaped as shown in Figure produces a magnetic field at P, the center of the arc. If the arc subtends an angle of 30.0° and the radius of the arc is 0.600 m, what are the magnitude and direction of the field produced at P if the current is 3.00 A?
Answer: 261 nT, into the page
Problem • A wire carrying a current of 5.00 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at the center of the loop is 10.0 μT, what is the required radius? • Answer: 31.4 cm
Problem • A cylindrical conductor of radius R = 2.50 cm carries a current of I = 2.50 A along its length; this current is uniformly distributed throughout the cross-section of the conductor. (a) Calculate the magnetic field midway along the radius of the wire (that is, at r = R/2). (b) Find the distance beyond the surface of the conductor at which the magnitude of the magnetic field has the same value as the magnitude of the field at r = R/2. • Answer: (a) 10.0 μT; (b) 2.50 cm
Problem • A uniform ring with a radius of 2.00 cm and a total charge of 6.00 μC rotates with a constant angular speed of 4.00 rad/s around an axis perpendicular to the plane of the ring and passing through its center. What is the magnetic moment of the rotating ring? • Answer: 4.80 × 10–9 A ∙ m2
Continuity Equation We know that the current density j and current I are related as
I =
∫∫
→
→
j • dS
S
The magnitude and direction of current are same everywhere and remain unchanged with time for the steady current. But, if the current is not steady, both current and current density vary from point to point as well as from time to time.
Thus we may express current as well as current density as → →
I ≡ I ( x, y , z , t )
and
j ≡ j ( x, y , z , t )
Let V be a volume of the current carrying conductor and ρ be the volume charge density, then total charge in the volume V is given by
q = ∫∫ ρ dV V
Since current is defined as the rate of flow of charge, the current through the conducting cylinder is
dq d I=− = − ∫∫∫ ρ dV dt dt V Where -ve sign signifies the outflow of charges from the cylinderical surfaces. Also
I=
→
→
∫∫ j • d S
Therefore, from last two equations we have
∫∫
d j • d S = − ∫∫∫ ρ dV dt V
→
→
Also according to the Gauss's divergence theorem →
→
→
∫∫ j • d S = ∫∫∫ div • j dV V
From the above two equations, we get d div • j dV = − ∫∫∫ ρ dV = ∫∫∫ dt V V →
dρ − dV ∫∫∫ dt V
dρ Div j = − dt →
Is known as equation of continuity and represents the physical fact of conservation of charge.
The divergence of the electric field gives a measure of the density of sources.
.
It also has implications for the conservation of charge
QUICK QUIZ 1
Consider the electric fields E1 at a distance r1 and E2 at a distance r2 from the center of a uniformly charged insulating sphere of total positive charge Q and radius a. If the radius of the sphere were suddenly doubled and the total charge Q were now uniformly spread throughout this new volume, the electric fields at the same locations, r1 and r2, would be a) E1 and E2, b) 2E1 and E2, c) E1/2 and E2, d) E1/8 and E2, e) E1/2 and 2E2, or f) E1/8 and 8E2.
Answer (d) The electric field inside the sphere is proportional to the charge density and the distance from the center of the sphere. The distance does not change but the charge density is reduced by a factor of 8 (since the volume increases by a factor of 8) when the radius increases by a factor of 2. Outside the sphere, since the total charge has not changed and the distance has not changed, Gauss’s law shows that the electric field must remain the same.
QUICK QUIZ 2 A conducting spherical shell (below) is concentric with a solid conducting sphere. Initially, each conductor carries zero net charge. A charge of +2Q is placed on the inner surface of the spherical shell. After equilibrium is achieved, the charges on the surface of the solid sphere, q1, the inner surface of the spherical shell, q2, and the outer surface of the spherical shell, q3, are a) q1 = Q, q2 = +Q, q3 = +Q b) q1 = 0, q2 = 0, q3 = +2Q c) q1 = 0, q2 = +2Q, q3 = 0 d) q1 = +Q, q2 = Q, q3 = +3Q
Answer (b) •
(b). Since the electric field inside the spherical conducting shell is zero, the net charge contained in a gaussian surface inside the shell that encloses the charges q1 and q2 will be zero. Since the solid sphere had initially zero net charge, from charge conservation, it will remain with zero net charge. Therefore, q1 must be zero, and since q1 + q2 = 0, q2 must also equal zero. The +2Q excess charge on the spherical shell will migrate to the outer surface of the shell. This is consistent with the fact that like charges repel and will spread as far apart from one another as possible.
Quick Quiz 3 Suppose the radius of a sphere (radius 1.00 m, with a charge of +1.00 μC at its center) is changed to 0.500 m. What happens to the flux through the sphere and the magnitude of the electric field at the surface of the sphere? (a) The flux and field both increase. (b) The flux and field both decrease. (c) The flux increases and the field decreases. (d) The flux decreases and the field increases. (e) The flux remains the same and the field increases. (f) The flux decreases and the field remains the same.
Answer: (e). The same number of field lines pass through a sphere of any size. Because points on the surface of the sphere are closer to the charge, the field is stronger.
Quick Quiz 4. In a charge-free region of space, a closed container is placed in an electric field. A requirement for the total electric flux through the surface of the container to be zero is that (a) the field must be uniform (b) the container must be symmetric (c) the container must be oriented in a certain way (d) The requirement does not exist – the total electric flux is zero no matter what.
Answer: (d). All field lines that enter the container also leave the container so that the total flux is zero, regardless of the nature of the field or the container.
Quick Quiz 5 If the net flux through a gaussian surface is zero, the following four statements could be true. Which of the statements must be true? (a) There are no charges inside the surface. (b) The net charge inside the surface is zero. (c) The electric field is zero everywhere on the surface. (d) The number of electric field lines entering the surface equals the number leaving the surface.
Answer: (b) and (d). Statement (a) is not necessarily true because an equal number of positive and negative charges could be present inside the surface. Statement (c) is not necessarily true, a nonzero electric field exists everywhere on the surface, but the charge is not enclosed within the surface; thus, the net flux is zero.
Quick Quiz 6 Consider the charge distribution shown in the figure. The charges contributing to the total electric flux through surface S’ are (a) q1 only (b) q4 only (c) q2 and q3 (d) all four charges (e) none of the charges
Answer: (c). The charges q1 and q4 are outside the surface and contribute zero net flux through S’.
Quick Quiz 7 Again consider the charge distribution shown in this figure. The charges contributing to the total electric field at a chosen point on the surface S’ are (a) q1 only (b) q4 only (c) q2 and q3 (d) all four charges (e) none of the charges
Answer: (d). We don't need the surfaces to realize that any given point in space will experience an electric field due to all local source charges
Problem1 • A spherical shell is placed in a uniform electric field. Find the total electric flux through the shell. • Answer: zero
Problem 2 • (a) A point charge q is located a distance d from an infinite plane. Determine the electric flux through the plane due to the point charge. (b) A point charge q is located a very small distance from the center of a very large square on the line perpendicular to the square and going through its center. Determine the approximate electric flux through the square due to the point charge. (c) Explain why the answers to parts (a) and (b) are identical. • Answer: (a) q/2ε0; (b) q/2ε0; (c) The plane and the square look the same to the charge.
Gauss' Law for Magnetism The net magnetic flux out of any closed surface is zero. This amounts to a statement about the sources of magnetic field. For a magnetic dipole, any closed surface the magnetic flux directed inward toward the south pole will equal the flux outward from the north pole. The net flux will always be zero for dipole sources. If there were a magnetic monopole source, this would give a non-zero area integral. The divergence of a vector field is proportional to the point source density, so the form of Gauss' law for magnetic fields is then a statement that there are no magnetic monopoles.
Problem 3 The following charges are located inside a submarine: 5.00μC, -9.00 μC, 27.0 μC, and -84.0 μC. (a) Calculate the net electric flux through the hull of the submarine. (b) Is the number of electric field lines leaving the submarine greater than, equal to, or less than the number entering it?
Problem 4. The electric field everywhere on the surface of a thin spherical shell of radius 0.750m is measured to be 890N/C and points radially toward the center of the sphere. (a) What is the net charge within the sphere’s surfaces? (b) What can you conclude about the nature and distribution of the charge inside the spherical shell?
Faraday's Law of Induction The line integral of the electric field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop. This line integral is equal to the generated voltage or emf in the loop, so Faraday's law is the basis for electric generators. It also forms the basis for inductors and transformers.
QUIZ For a long solenoid with closely spaced turns, each coil will a) exert an attractive force on the next adjacent coil, b) exert a repulsive force on the next adjacent coil, c) exert zero force on the next adjacent coil, or d) exert an attractive or repulsive force on the next adjacent coil, depending on the current direction in the solenoid.
(a). Because the coils are closely spaced, the distance between adjacent coils will be much smaller than the radius, r, of each coil and the forces between the coils can be approximated as the same as the force between two parallel wires of length 2pr. Since the currents are in the same direction, the force will be attractive.
Ampere's Law In the case of static electric field, the line integral of the magnetic field around a closed loop is proportional to the electric current flowing through the loop. This is useful for the calculation of magnetic field for simple geometries.
Q. A wire carrying a current of 5.00 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at the center of the loop is 10.0 μT, what is the required radius? Answer: 31.4 cm
Q. A cylindrical conductor of radius R = 2.50 cm carries • a current of I = 2.50 A along its length; this current is • uniformly distributed throughout the cross-section • of the conductor. • (a) Calculate the magnetic field midway along the radius of the wire (that is, at r = R/2). (b) Find the distance beyond the surface of the conductor at which the magnitude of the magnetic field has the same value as the magnitude of the field at r = R/2. Answer: (a) 10.0 μT; (b) 2.50 cm
Q. A current path shaped as shown in Figure produces a magnetic field at P, the center of the arc. If the arc subtends an angle of 30.0° and the radius of the arc is 0.600 m, what are the magnitude and direction of the field produced at P if the current is 3.00 A?
Answer: 261 nT, into the page
Problem • A wire carrying a current of 5.00 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at the center of the loop is 10.0 μT, what is the required radius? • Answer: 31.4 cm
Problem • A cylindrical conductor of radius R = 2.50 cm carries a current of I = 2.50 A along its length; this current is uniformly distributed throughout the cross-section of the conductor. (a) Calculate the magnetic field midway along the radius of the wire (that is, at r = R/2). (b) Find the distance beyond the surface of the conductor at which the magnitude of the magnetic field has the same value as the magnitude of the field at r = R/2. • Answer: (a) 10.0 μT; (b) 2.50 cm
Problem • A uniform ring with a radius of 2.00 cm and a total charge of 6.00 μC rotates with a constant angular speed of 4.00 rad/s around an axis perpendicular to the plane of the ring and passing through its center. What is the magnetic moment of the rotating ring? • Answer: 4.80 × 10–9 A ∙ m2
Continuity Equation We know that the current density j and current I are related as
I =
∫∫
→
→
j • dS
S
The magnitude and direction of current are same everywhere and remain unchanged with time for the steady current. But, if the current is not steady, both current and current density vary from point to point as well as from time to time.
Thus we may express current as well as current density as → →
I ≡ I ( x, y , z , t )
and
j ≡ j ( x, y , z , t )
Let V be a volume of the current carrying conductor and ρ be the volume charge density, then total charge in the volume V is given by
q = ∫∫ ρ dV V
Since current is defined as the rate of flow of charge, the current through the conducting cylinder is
dq d I=− = − ∫∫∫ ρ dV dt dt V Where -ve sign signifies the outflow of charges from the cylinderical surfaces. Also
I=
→
→
∫∫ j • d S
Therefore, from last two equations we have
∫∫
d j • d S = − ∫∫∫ ρ dV dt V
→
→
Also according to the Gauss's divergence theorem →
→
→
∫∫ j • d S = ∫∫∫ div • j dV V
From the above two equations, we get d div • j dV = − ∫∫∫ ρ dV = ∫∫∫ dt V V →
dρ − dV ∫∫∫ dt V
dρ Div j = − dt →
Is known as equation of continuity and represents the physical fact of conservation of charge.
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