«Notes and Solved Problems for Common Exam 2 3. GAUSS LAW Key concepts: Gaussian Surface, Flux, Enclosed Charge Gauss Law is equivalent to the Coulomb ...»
Notes and Solved Problems for Common Exam 2
3. GAUSS LAW
Key concepts: Gaussian Surface, Flux, Enclosed Charge
Gauss Law is equivalent to the Coulomb law but sometimes more useful. Gauss
law considers a flux of an electric field thru some hypothetical surface (called a
Gaussian surface) The flux of the field E thru the surface S is formally an integral
Φ = ∫ EdS
The Gauss law relates the flux of the electric field thru the Gaussian surface with
the total charge enclosed by this surface rr q Φ = ∫ EdS = total ε0 The usefulness of the Gauss law is seen when calculating electric fields near symmetrical objects. For these situations, the electric field can for example be a constant on the surface of the integration and can be taken out of the integral defined above. The integral can then often be done easily (it is just the area of the Gaussian surface) and one can immediately find and expression for the electric field on the surface.
2 The units of electric flux: Nm / C Gauss’s Law helps us understand the behavior of electric fields inside the conductors. Since the conductors are the objects where the electrons move freely the electric field must be zero everywhere inside the conductor. The reason: if a conducting object is placed in a non-zero external electric field the charge inside the conductor will move towards its surface until the electric field created by this displaced charge completely compensates the external electric field. The total electric field inside the conductor is therefore zero. If that were not so, free charges would move due to the field and make it so.
The Gauss law also helps us understand the distribution of electric charge placed into a conductor. Since the electric field inside every conductor is always zero, there cannot be extra electric charge located inside of one. Therefore, any extra charge placed into the conductor will move until it is distributed only on the surface of the conductor. This can be proven by considering any Gaussian surface lying completely inside the conductor, since electric field inside the conductor is zero, there is no charge which is enclosed inside this Gaussian surface.
Typical problems related to Gauss Law:
Problem 12. Find the flux through a spherical Gaussian surface of radius a = 1 m surrounding a charge of 8.85 pC.
A) 1 x 10-16N m2/C B) 1 x 10-12 N m2/C C) 1 x 10-8 N m2/C D) 1 x 10-4 N m2/C E) 1 N m2/C Solution. The flux thru the Gaussian surface is the charge located inside the −12 −12 2 surface. Therefore: Φ = q / ε 0 = 8.85*10 / 8.85*10 = 1Nm / C Answer E Problem 13. A positive charge Q= 8 mC is placed inside the cavity of a neutral spherical conducting shell with an innerradius a and an outer radius b. Find the charges induced at the inner and outer surfaces of the shell.
A) Inner charge = –8 mC, Outer charge = +8 mC
B) Inner charge = +8 mC, Outer charge = -8 mC
C) Inner charge = 0 mC, Outer charge = +8 mC
D) Inner charge = –8 mC, Outer charge = 0 mC
E) Inner charge = 0 mC, Outer charge = 0 mC Solution. Inside any conductor the electric field must equal zero. Therefore, the electric field created by the external charge inside the shell is cancelled by electric fields created by the induced charged on the inner and outer surfaces of the shell. Choosing the Gaussian surface to enclose the external charge and the Φ = ∫ EdS = 0 = Q + qin
inner surface of the shell and applying the Gauss law:
since the electric field inside the conductor is zero. This gives us qin = −Q = −8mC Since the conducting shell is electrically neutral, the charge induced on the inner part of the shell has been taken from the outer surface of the shell which implies qin + qout = 0 = qout = 8mC Answer A.
Problem 14. A positive charge Q=8 mC is placed inside a spherical conducting shell with inner radius a and outer radius b which has an extra charge of 4 mC placed somewhere on it. When all motion of charges ends (after 10-15 sec), find the charges on the inner and outer surfaces of the shell.
A) Inner charge = –8 mC, Outer charge = +8 mC
B) Inner charge = +8 mC, Outer charge = -8 mC
C) Inner charge = +8 mC, Outer charge = -12 mC
D) Inner charge = –8 mC, Outer charge = 12 mC
E) Inner charge = -4 mC, Outer charge = +4 mC Solution. Inside any conductor the electric field equals zero. Therefore, the electric field created by the external charge inside the shell is compensated by the electric fields created by the induced charges on the inner and outer surfaces of the shell. Choosing the Gaussian surface to enclose the external charge and Φ = ∫ EdS = 0 = Q + qin
the inner surface of the shell and applying the Gauss law:
since the electric field inside the conductor is zero. This gives us Q = − qin = −8mC The conducting shell here is not electrically neutral, but has some extra charge 4 mC. Since net charge has to remain constant (none of it can escape), the charge on the outer surface of the shell equals the net charge plus a charge equal in magnitude to that on the inner surface but of the opposite sign. The condition is qin + qout = 4mC = qout = 12mC Answer D.
Problem 15. Find the value of the electric field at a distance r= 10 cm from the center of a non conducting sphere of radius R= 1 cm which has an extra positive charge equal to 7 C uniformly distributed within the volume of the sphere.
Problem 16. A positive charge is placed inside a spherical metallic shell with inner radius a and outer radius b. The charge is placed at shifted position relative to the center of the shell. Describe the charge distribution induced at the shell surfaces.
A) A negative charge with uniform surface density will be induced on the inner surface, a positive charge will be induced on the outer surface.
B) A negative charge with non-uniform surface density will be induced on the inner surface, a positive charge will be induced on the outer surface.
C) A positive charge with uniform surface density will be induced on the inner surface, a negative charge will be induced on the outer surface.
D) A positive charge with non-uniform surface density will be induced on the inner surface, a negative charge will be induced at the outer surface.
E) A negative charge with uniform surface density will be induced on the inner surface, a positive charge will be induced on the outer surface.
Solution. First, since the external charge placed inside has a positive sign, the inner surface will be charged negatively, and the outer surface will be charged positively. There is no charge induced in the bulk of the metallic shell. Second since the external charge is not precisely at the center of the shell, the induced charge density on the inner surface will not be uniform.
4. ELECTRIC POTENTIAL
Just as we visualize charged objects creating an electric field around them, they also create a potential field. The potential field or equivalently electric potential is
directly related to the electric field:
V ( x ) = − ∫ E ( x )dx
In some problems the concept of electric potential may be more useful than the concept of electric field since the potential is a scalar function (scalar defined for each point of the space) while the electric field is a vector function (vector defined for each point of the space)
The potential can be visualized by drawing so called equipotential surfaces (surfaces of the same potential). For example, for the point charge, these surfaces are the spheres centered at r =0 since at all points of the surface, r is q the same, therefore V ( r ) = k is the same, which produces an equipotential r surface. For the potential created by the infinite plate, the equipotential surfaces are the planes parallel to the charged plate since the potential is V ( x ) = − Ex + const. and all points separated by the distance x from the plate form a plane surface.
Note that the equipotential surfaces are always perpendicular to the electric field lines. For example, the electric field lines for a point charge are lines originating radially from the origin, while equipotential surfaces are spheres centered at the origin. For an infinite charged plate, the electric field lines are the lines perpendicular to the plate while equipotential surfaces are the planes parallel to the plate.
Note also that it costs zero energy to transfer charge between any two points on an equipotential surface: Since the work done by the electric field described by the potential V(r) to move the charge Q from the point r1 to the point r2 is given
W = Q *[V ( r1 ) − V ( r2 )], this work is always equal zero as long as V ( r1 ) = V ( r2 ) which is the formal definition of an equipotential surface.
Therefore, the positively charged particle will always move from the region of the higher potential to the region of the lower potential. Likewise, a ball placed on top of a hill will move down the hill since the potential energy of the ball on top of the hill is larger than its potential energy lower down, and the ball “wants” to minimize its potential energy. (Here we should recall the relationship for the gravitational potential V (h ) = mgh, where m is the mass of the ball, g is free fall acceleration and h is the height.) For negatively charged particles the situation is reversed: negatively charged particles would always move from the region of the lower potential to the region of the higher potential, although they are still moving spontaneously from higher to lower potential energy. The example of the ball does not represent this: a ball placed near the bottom of a hill will not spontaneously move upward. This does not happen of course since there is no such thing as negative masses in mechanics.
Typical problems related to electric potential:
Problem 1. Calculate the difference between the potential at r = 5 cm and r = 10 cm for a single point charge of 1 C located at the origin.
Problem 2. What are the equipotential surfaces of the uniformly charged sphere?
A. Equipotential surfaces are the planes crossing the sphere and perpendicular to z axis B. Equipotential surfaces are the cylinders with the common axis along z direction C. Equipotential surfaces are the spheres centered at the center of the charged sphere D. Equipotential surfaces are elliptical and stretched along z direction.
E. Insufficient information.
Solution. For the uniformly charged sphere the electric potential at r larger than the sphere radius is the same as for the point charge concentrated at the center of the sphere. This is the same as for the electric field created by the uniformly charged sphere, which can be found as from the same point charge placed at the sphere center. Therefore the equipotential surfaces are spheres centered at the origin.
Problem 3. Find the electric potential at the center of the equilateral triangle whose side is 1m if there are three positive charges of 1 C, 2 C and 3 C in its corners.
(Assume that V(r)=0 when r goes to infinity) A. 9.07*10-10 V B. 3.21*10-5 V C. 7.09*100 V D. 9.23*10+10 V E. 2.31*10+14 V Solution. The electric potential at the center of the triangle is the sum of the electric potential from each charge: V = V1 ( r ) + V2 ( r ) + V3 ( r ) where r is the distance between the charge and the center of the triangle which is the same for all three charges (as long as the triangle is equilateral). If the side is 1 m, the distance to the center is 0.5/cos(30)=1/ 3. We finally have V = kq1 / r + kq2 / r + kq3 / r = 9 *109 * 3 * (1 + 2 + 3) = 9.23*1010V Answer D.
Problem 4. If the potential has a form of stairs at which points the electric field is the largest?
A. At the bottom of the stairs.
B. At the top of the stairs C. At flat area of each step of the stairs D. At transition points between the steps E. The electric field is constant across the stairs.
Solution. The electric field is the derivative of the potential with respect to x.
Therefore the electric field is the largest at the points where the derivative of the potential with respect to x is largest. If the potential has a step like form, its derivative is zero at all flat areas of the steps while the derivative goes to infinity at transition points between each step (since the derivative is simply given by the slope, i.e. tan(0)=infinty). Therefore electric field is infinite at transition points and zero everywhere else.
Problem 5. Find the value of the surface charge density for the infinitely charged plate if it is known that each two equpotential surfaces separated by 1 m have potential difference of 5 V.
A. 8.85*10-11 V B. 8.85*10-9 V C. 8.85*10-7 V D. 8.85*10-5 V E. 8.85*10-3 V Solution. For infinitely charged plate, the electric field is constant and it is given σ
by E =. The potential corresponding to the constant field is given by:
2ε 0 V ( x ) = − Ex + const. The equipotential surfaces between any two points x1 and x2 have the potential difference: V ( x1 ) − V ( x2 ) = − E ( x1 − x2 ). From here we conclude that the absolute value of the electric field E = (V ( x1 ) − V ( x2 )) /( x1 − x2 ) = 5/1 = 5V / m = 5N / C It follows that σ = 2 Eε 0 = 2 *5*8.85*10−12 = 8.85*10−11 C / m 2 Answer A.
A typical capacitor consists of metallic plates separated by some distance.
If a capacitor is charged (the plates have charge +Q and –Q), it maintains a potential difference between the plates.
If the plates have a charge +Q and –Q, the charge is related to the potential difference between the plates by
where C is called the capacitance of the capacitor. If a battery with potential difference V is attached to capacitor, the charge after a long time will be Q=C*V Units of C are “Farads”: 1 F= 1 Coulomb/1 Volt.
Frequently used units also are µF=10-6 F and pF = 10-12 F.