Physics Final Exam Review: Test One Material

C (when the area and electric field are perpendicular the flux is 0, which is what occurs at point A, and only C indicates this; at B the flux should be the area, which is 6 multiplied by the electric field which is 12 N/C, so the flux at point B is 72)

A box is immersed in a uniform electric field pointing to the right as shown. What is the electric flux through surface (facing you) A and surface B? A) 220 0 B) -220 -72 C) 0 +72 D) 220 +72 E) 140 0

D (capacitance is a property of the physical system, and does not vary with applied voltage; if the voltage is doubled, the charge will double)

A capacitor stores charge Q at a potential difference ∆V. What happens to the capacitance, C, if the voltage applied to the capacitor by a battery is doubled to 2∆V? A) the capacitance falls to half its initial value, and the charge remains the same B) the capacitance and the charge both fall to half their initial values C) the capacitance and the charge both double D) the capacitance remains the same, and the charge doubles

C (work is equal to the charge multiplied by the change in potential difference, in this case the charge will be the same, so work will be dependent on the value of ∆V; here A--> B will be 0 since it is not changing levels, and D--> E will be -1, giving it the lowest ∆V value)

A diagram of "equipotential surfaces." Rnak (from largest to smallest) the work done by the electric field on a positively charged particle that moves from A--> B; from B-->C; C--> D; and from D--> E. A) A-->B; B--> C; C--> D; D--> E B) B--> C; C--> D; D--> E; A--> B C) B--> C; C--> D; A--> B; D--> E

E (work is independent of the path taken, only depends on final and initial points)

A large negative charge -Q is located in the vicinity of points A and B. Suppose a positive charge +q is moved at constant speed from A to B by an external agent. Along which of the paths shown in the figure will the work done by the electric field be the greatest? A) Path 1 B) Path 2 C) Path 3 D) Path 4 E) work is the same along all four paths

A (fluxes cancel on opposite sides of the spheres)

A point charge +Q is located on the x axis at x=a, and a second point charge -Q is located on the x-axis at x=-a. A Gaussian surface with radius r=2a is centered at the origin. First make a sketch ( or just visualize) the E field lines you epxect here. Then find that the electric flux through this Gaussian surface is A) zero because the negative flux over one hemisphere is equal to the positive flux over the other 2. greater than zero 3. zero because at every point on the surface the electric field has no component perpendicular of the surface 4. zero because the electric field is zero at every point on the surface 5. none of the above

(image shown in figure attached: note that the charged is located at a distance of half the radius and the electric field lines should be PERPENDICULAR to the conductor both inside and outside)

A positively charged particle is at a distance R/2 from the center of an uncharged thin, conducting, spherical shell of radius R. Sketch the electric field lines set up by this arrangement both inside and outside the shell.

C (by induction before contact, electrons move away from negatively charged rod, making the ball positively charged on surface closest to the rod, so it will attract the rod; after contact some of the negative charge will transfer to the sphere, and the excess negative charge will cause it to repel the rod)

A small pith ball is hanging by a string. I bring a negatively charged rod next to it, and slowly move it towards the ball. I then touch the two. What happens? A) before contact the ball is repelled by the rod, and after contact the ball is repelled by the rod B) before contact the ball is repelled by the rod; after contact the ball is attracted to the rod C) before contact the ball is attracted to the rod; after contact the ball is repelled by the rod D) before contact the ball is attracted to the rod; after contact the ball is attracted to the rod

A (at A, we use the following value for q, which comes from q = rho*volume: q*4/3*π*r^3 divided by 4/3*π*R^3, which gives q*r^3/R^3, where r=4cm, R = 5cm) (final answer for part A: E*4πr^2 = (4/5)^3*q/epsilon not ---> E = (4/5)^3*q/(4πr^2*epsilon not)) (for B, the radius is in between 5 cm and 10 cm, and the charge enclosed is the charge of the insulating sphere, which here is just q; here E = q/(4π8^2*epsilon not)) (because C is located within a conductor its value is zero) (here the q enclosed is the net charge of the conducting ring and the insulating sphere, which again is just q in this case; here the E value is calculated the same way as it would be in part B, but its value will be larger because the radius is much larger: E = q/(4π*16^2*epsilon not))

A solid insulating sphere of radius 5 cm carries an electric charge "Q" uniformly distributed throughout its volume. Concentric with the sphere is a conducting spherical shell with no net charge as shown in Figure below. The inner radius of the shell is 10 cm and the outer radius is 15 cm. No other charges are nearby. Rank the magnitude of the electric field at points A (at radius 4 cm), B (radius 8 cm), C (radius 12 cm) , and D (radius 16 cm) from largest to smallest. Display any cases of equality in your ranking. A) A>B>D>C B) B=D>A>C C) C>B=D>A D) C>B>A=D E) none of these

(A: qin = rho*4/3*π*r^3 = Q*4/3*π*r^3 divided by 4/3*π*R^3 = Q*[r/R]^3) (B: E = k*Qr/a^3) (C: qin = Q; D: E = k*Q/r^2) (E: 0 because there is no electric field inside a conductor) (F: E = 0, where E = [Q+Qinner]/epsilon not, so Q+Q inner = 0; therefore Q inner = -Q) (G: Qouter = -Q inner = +Q) (H: A surface of area A holding charge Q has a surface charge of sigma = q/A; the solid, insulating sphere has small surface charge because its total charge Q is uniformly distributed throughout its volume; the INNER SURFACE OF RADIUS b has a smaller surface area, and therefore a LARGER SURFACE CHARGE, than the outer surface area of radius c)

A solid, insulating sphere of radius a has a uniform charge density throughout its volume and a total charge Q. Concentric with this sphere is an uncharged, conducting, hollow sphere whose inner and outer radii are b and c as shown in the figure. We wish to understand completely the charges and electric fields at all locations A) Find the charge contained within a sphere of radius r<a. B) From this value, find the magnitude of the electric field for r<a C) What charge is contained within a sphere of radius r when a<r<b? D) From this value, find the magnitude of the electric field for r when a<r<b. E) Now consider r when b<r<c. What is the magnitude of the electric field for this range of values of r? F) From this value, what must be the charge on the inner surface of the hollow sphere? G) From part (f), what must be the charge on the outer surface of the hollow sphere? H) Consider the three spherical surfaces of radii a, b, and c. Which of these surfaces has the largest magnitude of surface charge density.

(for part A assuming that the rod is a cylinder and that you are using the rounded surface so area = 2πrl) (A: E*2πrl = lambda*l/epsilon no; ANSWER: E = 2*lambda*k/r) (B: forces acting on the electron are gravity and electric force, so Fe - mg = 0, and Fe = mg = 9.11x10^-31*9.81 = 8.93x10^-30 N) (C: final answer should be 3.7x10^-23 C/m; Fe = q*E, where q= 1.6x10^-19, and E is equal to what was calculated above, and Fe is the answer from part B)

An electron is suspended at a distance of 1.2 cm above a long uniform line of charge linear charge density lambda, is unknown. Reminder: Earth's gravity is actin gon the electron as well. A) Using Gauss's Law, determine an expression for the electric field due to the line of charge in terms of lambda at the position of the electron. B) Sketch the diagram with the forces actingin on the electron, determine the electric force on the electron. C) Using your answers to parts (a) and (b) above, determine the linear charge density for this case.

(A: q enclosed = 6nC*(r1/2 / r1)^3 = 6x10^-6 * 1/8) (E*4π*r1^2 = 6x10^-6*1/8/epsilon not; E = 0.75nC/[4πepsilon not]) (B: Q enclosed = 12 -6 = 6 nC, and radius is r3; E = 6nC/r3^2 * 1/4π*epsilon not)

An insulating sphere with net charge of -6 nC uniformly distributed throughout its voume is placed at the center of a conducting spherical shell as shown in the cross section figure. A) If the net charge on conducting shell is +12 nC, use Gauss's law to determine the electric field at r1/2. B) Use gauss's law to determine the electric field just outside the outer surface of the shell (at r3)

constant (VB-VA = 0, so they are the same at both points)

Because the electric field is zero inside a conductor, we conclude that the electric potential is ______ everywhere inside the conductor and equal to the value at the surface.

rho = Q/V (units coulomb/m^3); sigma = Q/A (units coulomb/m^2); lambda = Q/l (units coulomb/m) dq = rho*dV; dq = sigma*dA; dq = lambda*dl

Charges distributed evenly vs. nonuniformerly: Uniform volume charge density = ______ surface charge density = _______ linear charge density = _______ Nonuniform: volume: _______ surface area: _______ length: _______

(A: C = Q/∆V = Q/Ed, where E = sigma/epsilon not; note that sigma = Q/A, so E = Q/A/epsilon not = Q/A*epsilon not) (plugging this into the capacitance equation gives C = Q/Q/epsilon not*A *d = epsilon not *A/d) (B: Q = C∆V = epsilon not*A/d * ∆V = 8.6x10^-9 C) (C: E = ∆V/d = 9V/0.025x10^-3 = 3.6x10^5 N/C or V/m)

Consider a parallel plate capacitor where the electric field is constant with each plate having area of 27cm^2 and separation of 0.025mm. A) Find an expression for the capacitance by starting with the definition, C = Q/∆V. B) If this capacitor is connected to a 9-volt battery, find the charge, Q, on the capacitor. C) Using your results from above, determine the electric field between the plates.

(B: V = E*d = 500*0.0045 = 22.5 V) (C: U = q*∆V = 1.6x10^-19*22.5 = 3.6x10^-18 J, this is set equal to the kinetic energy equation, 0.5mv^2, solving for velocity; the mass of a proton is 1.67x10^-27 kg; the final answer equals 6.5x10^4 m/s)

Consider a proton at point a in a uniform electric field (E=500 N/C). A) Draw a label an electric field line in the sketch. (see figure) B) find the change in potential when moving from point a to point b C) Find the velocity of the proton when allowed to move to point b under the influence of the electric field. Ignore gravity in this case.

(V = ke∫dq/r = k*Q/√a^2+x^2; and Ex = k*x*Q/[a^2+x^2]^3/2) (∆V = V of 2R - V of 0 = k*Q/√R^2 + 4R^2 - k*Q/R = k*Q/R*(1/√5 - 1) = -0.553*k*Q/R)

Consider a ring of radius R with a total charge Q spread uniformly over its perimeter. What is the potential difference between the point at the center of the ring and a point on its axis a distance 2R from the center? *uniformly charged RING

(A: both spheres must be at the same potential according to k*q1/r1 = k*q2/r2, where q1+q2 = 1.2x10^-6 C; q1 also equals q2*r1/r2) (Therefore: q1 + q2 because (q2*r1/r2)+q2 = 1.2x10^-6, so when we solve for q2 we get 0.3x10^-6 C, and when we solve for q1 we get 0.9x10^-6 C, which we then plug into V = ke*q1/r1) (The potential difference equals 1.35x10^5 V) (B: outside the larger sphere E1 = k*q1/r1^2 = V1/r1 = 2.25*10^6 V/m away; outside the smaller sphere E2 = V/r2 = 6.74x10^6 V/m away; the smaller sphere carries less charge but creates a much stronger electric field than the larger sphere)

Electric charge can accumulate on an airplane in flight. You may have observed needle-shaped metal extensions on the wing tips and tail of an airplane. Their purpose is to allow charge to leak off before much of it accumulates. The electric field around the needle is much large than the field around the body of the airplane and can become large enough to produce dielectric breakdown of the air, discharging the airplane. To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20 microC charge is placed on the combination. One sphere, representing the body of the airplane, has a radius of 6.00 cm; the other, representing the tip of the needle, has a radius of 2.00 cm. (A) What is the electric potential of each sphere? (B) What is the electric field at the surface of each sphere?

(A: electric field at A is greater than at B because E = ∆V/∆s, and since the distance is smaller, the field is greater) (B: electric field at B = -∆V/∆s = -(6-2)/2cm = 200 N/C down) (C: shown in figure = original figure did not have the field lines)

Figure shows several equipotential lines, each labeled by its potential in volts. The distance between the lines of the square grid represents 1.00 cm. (A) Is the magnitude of the field larger at A or at B? Explain how you can tell. (B) Explain what you can determine about E at B. (C) Represent what the electric field looks like by drawing at least eight field lines.

flux = 339 N*m^2/C (No the electric field is not uniform on this surface; Gauss's law is only practical to use when all portions of the surface satisfy one or more of the conditions listed)

Find the net electric flux through the cube in the figure shown. Can you use Gauss's law to find the electric field on the surface of this cube? Explain.

-226 N*m^2/C (flux = charge enclosed/epsilon not = -2nC/8.85e-12)

Find the net electric flux through the spherical closed surface shown in the figure. The two charges on the right are inside the spherical surface.

A (the forces along OB and OC cancel each other; there is one force along OD, and then a force pointing up and another point to the left, whose results will be greater than each individual force and pointed along OA, making the final resultant force pointing along OA)

Five equal charges +1 are placed on a square as illustrated. A fifth charge +q' is placed at the cneter O. What is the direction of the resultant force on q'? A) along OA B) along OB C) along OC D) along OD E) along some other direction F) need more information

toward; attracted

For a negative source charge, the electric fields lines are directed ______ the source charge, because the positive test charge would be _______ toward the negative source charge.

(E*A = qenclosed/epsilon not, where area = 4πr^2; eventually get the equation E = k*q/r^2) (applies to symmetric shapes, such as spheres only)

Gauss's law in symmetric circumstances: Look at figure, and how should Gauss's law be applied?

(A: V = k*q/r, so Va = 8990V and Vb = 8960 V) (B: E = ∆V/∆r = 8990-8960/3.01-3 = 3000 V/m or N/C)

Given a +3 microC point charge and r is the direction and distance from the charge. A) Find the electric potential at ra = 3.00 m and at rb = 3.01 m. B) Using your answers from part a and the relationship between the electric field and electric potential, find the electric field at 4=3.00m.

E (because all of them have areas parallel to the field, there is an electric field at all of these points that can be determined using the flux)

Given a very large sheet of uniformly distributed charge, which of the following Gaussian surfaces can be used to determine the electric field near the sheet of charge? A) A cube with two faces parallel to the sheet B) a rectangular box with two faces parallel to the sheet C) a circular cylinder with the end faces parallel to the sheet D) a sphere centered on the sheet E) all of the above

(A: net force equals 0; F = q1*q2*k/r^2, which is the same and in opposite directions for both charges, so they will cancel) (B: electric force = q*E, so the electric field is also zero) (C: V = k*q/r, and there are two of the same charges so the potential differences = 2*kq/r = 45.0 kV) (how can E=0, but the potential difference does not: this is because E = -dV/dx which is the rate of change of V with position; V here is constant, so E is equal to 0)

Given two particles with 2.00-microC charges as shown in the figure and a particle with charge q=1.28x10^-18 C at the origin, A) What is the net force exerted by the two 2.00 microC charges on the charge q? B) What is the electric field at the origin due to the two 2.00-microC particles? C) What is the electric potential at the origin due to the two 2.00 microC particles?

B (electrons in sphere move away from negatively charged rod, and down the rod making both of the foils excessively negative, and thus they will repel each other)

I bring a negatively charged rod next to an electroscope without touching it. The electroscope is made of metal. What happens? A) nothing B) the two hanging foils move away from one another C) the two hanging foils move towards one another

nonuniform

In a _______ field the lines at different locations points in different directions.

A (in this case the potential is a constant value, so the derivative is zero)

In a certain region of space, the electric potential is +2V everywhere along the x-axis. From this information, you can conclude that the x-component of the electric field in this region is A) zero B) in the positive x direction C) in the negative x direction D) more information is needed

A (its derivative along this direction must be zero)

In a certain region of space, the electric potential is zero everywhere along the x-axis. From this information, you can conclude that the x-component of the electric field in this region is A) zero B) in the positive x direction C) in the negative x direction D) more information is needed

A (equation used is ∆V = qnot*∆V, so if a negative test charge is moved through a negative potential difference, the change in potential energy is positive; work must be done to move the charge in the direction opposite the electric force on it) (NOTE ∆V is negative by the following equation: ∆V = -∫E•ds, where E and ds are both positive, so this value is also negative and when multiplied by a negative charge, the potential energy is positive)

In the figure below, two points A and B are located within a region in which there is an electric field. A negative charge is placed at A and then moved to B. How would you describe the change in potential energy of the charge-field system for this process? A) it is positive B) it is negative C) it is zero D) need more information

C (the potential is established only by the source charge and is independent of the test charge)

In the figure, take q1 to be a negative source charge and q2 to be the test charge. If q2 is initially positive and is changed to a charge of the same magnitude but negative, what happens to the potential at the positive of q2 due to q1? A) it increases 2. it decreases 3. it remains the same 4. more information is needed

A (initially by the equation U = k*q1*q2/r12 is negative because the charges are opposite charges, however, if q2 is changed to a negative value, then they are both negative, and now the potential energy is positive, so it is increasing)

In the figure, take q1 to be a negative source charge and q2 to be the test charge. When q2 is changed from positive to negative, what happens to the potential energy for the two-charge system? A) it increases B) it decreases C) it remains the same D) more information is needed

B (when moving straight from A to B, E and ds both point toward the right, and the equation for the potential difference is ∆V = -∫E•ds; because both are positive, the product for ∆V is negative)

In the figure, two points A and B are located within a region in which there is an electric field. How would you describe the potential difference ∆V = VB-VA? A) it is positive B) it is negative C) it is zero D) need more information

(a cylindrical gaussian surface penetrating an infinite plane of charge; the flux is EA through each end of the gaussian surface and is zero through its curved surface) (E and dA should be parallel to one another, which is why the curved area is ignored because it is perpendicular to the field)

Note the figure. The electric flux in this case is equal to E*2A because you need to consider both of the surfaces of the cylinder.

1 (C = 0 because there are no lines there; A is in an area where the lines are the densest so it is the strongest)

Rank the magnitudes of the electric field at points A, B, and C shown in the figure from greatest to least. 1. A, B, C 2. C, A, B 3. B, C, A 4. C, B, A 5. A, C, B 6. B, A, C

(1: because q is inside the insulating sphere we need to use the following equation to note that q is uniformly distributed around the sphere: q*4/3*π*r^3 divided by 4/3*π*R^3, where r is the radius of the charge within the sphere, and R is the radius of the sphere) (1 continued: the area is equal to the surface area covered by the charge which is 4πr^2, where r is the radius of the charge within the sphere; overall the electric field by Gauss's law = [r/R]^3*q divided by area*epsilon not) (2: here the charge enclosed is the charge of the insulating sphere, or the q value given; the area is equal to 4πr^2, where r is the radius where the charge is located in between the conducting ring and the insulating sphere) (3: here because the charge is located within the conductor, the value of the electric field here is zero) (4: q enclosed will be the net charge of the conducting ring and the insulating sphere, the area will be 4πr^2, where r is the radius where the charge is located)

Solving Gaussian problems for conducting ring around an insulating sphere: 1) If the radius of the charge (little r) is less than the radius of the insulating sphere (big R): how do we find the electric field value? 2) if the radius of the charge (little r) is greater than that of the insulating sphere but less than that of the conducting ring. How do we find the electric field value? 3) if the radius is located within the conducting ring, how do we find the electric field value? 4) if the radius is located outside of the conducting sphere, how do we find the electric field value?

zero; on the surface; sigma/epsilon; sigma; sigma; radius of curvature

The electric field inside a conductor is equal to ________. All the charge must reside _______. The magnitude of the electric field just outside the charge conductor has a magnitude of ______, where ____ is the surface charge density at that point. For an irregularly shaped conductor, _____ is greatest where the ______ is smallest.

(A: when r<R, V = k*Q/R, and Er = -dV/dr = 0 because the charge is inside a conducting sphere) (B: when r>R, V = k*Q/r, and Er = -dV/dr = -(-k*Q/r^2) = k*Q/r^2, same as in earlier examples)

The electric potential inside a charged spherical conductor of radius R is given by V = kQ/R. The potential outside the sphere is V = kQ/r. Find the E field in these two cases: A) inside the sphere B) outside the sphere

(q1/q2 = -6/18 = -1/3) (q1 is negative because lines are pointing towards the charge; q2 is positive because lines are pointing outward from charge)

The figure shows the electric field lines for two charged particles separated by a small distance A) determine the ratio q1/q2 b) What are the signs of q1 and q2?

away; repelled

The lines are directed ______ from the source charge, when it is positive because a positive test charge would be ________ away from the positive source charge.

A (F1 = k*Q*-Q/a^2; F2 = k*2Q*Q/(2a)^2; total force = F1+F2)

What is the electric force acting on charge q3? A) -kQ^2/(2a^2) B) zero C) +kQ^2/(2a^2) D) +kQ^2/(a^2) E) none of these

(a = sketch should show same number of lines pointing outward from each charge) (b = at the center because there are no electric field lines that would go through this region, therefore there is no electric field here) (c: E = ke*q/r^2 or in this case for both charges E = ke*q/a^2) (angles are 60 degrees because it is an equilateral triangle, so E = E1 + E2 = ke*q/a^2 * [cos60 + sin60 - cos60 + sin 60], which equals ke*q/a^2 * (2*sin 60) = 1.73*ke*(q/a^2)j)

Three equal positive charges q are at the corners of an equilateral triangle of side a as shown in the figure. Assume the three charges together create an electric field. a) Sketch the file d lines in the plane of the charges. b) Find the location at one point other than infinity where the electric field is zero. c) What is the magnitude of the electric field at P due to the two charges at the base? d) What is the direction of the electric field at P due to the other two charges at the base?

(answer: magnitude = 1.384x10^-5; direction = 77.5 degrees below the -x axis)

Three point charges are arranged as shown in the figure. Fina (a) the magnitude and (b) the direction of the electric force on the particle at the origin.

A (because the two fields are identical, they should have the same number of lines at each charge; because the charges are negative, the field lines should be pointed towards both of the charges; also the lines should indicate repulsion since both of the charges are the same sign = this is only supported by diagram A)

Two identical negative point charges are shown along with possible electric fields caused by the two charges. identify the sketch with the correct electric field.

(NOTE: E = sigma/2*epsilon not, but in this case there are two slabs so E = sigma/epsilon not) (A: to the left of the slabs, the electric fields from both slabs point in opposite directions and cancel each other so the overall electric field magnitude is ZERO) (B: in the region between, both sheets point in the direction of the negative slab, so the electric field = sigma/epsilon not, pointing to the right) (C: as in part A, the electric fields point in opposite directions and therefore cancel, so the electric field is equal to ZERO) (D: if both are the same value then the electric fields to the left and right of each slab will be double what they were in the previous, because there will be two electric fields pointing in the same direction: E = 2*sigma/epsilon not; however in between there will be two electric fields pointing in opposite directions so the electric fields will cancel and the net electric field will be equal to ZERO) (drawing diagrams may be helpful here)

Two infinite, nonconducting sheets of charge are parallel to each other as shown in the Figure. The sheet on the left has a uniform surface charge density sigma, and the one on the right has a uniform charge density of -sigma. Calculate the electric field at points (a) to the left of, (b) in between, and (c) to the right of the two sheets. D) What if? Find the electric fields in all three regions if both sheets have positive uniform surface charge densities of value sigma.

(A: U total = k*(q1q2/r12 + q2q3/r23 + q1q3/r13 = -4.5x10^-5 J; note that the negative values for charges need to be accounted for in the equation above) (B: .5mv^2 initial + qV initial = 0.5mv^2 final + qV final, where qV final and kinetic energy initial are both 0; v = 3.46x10^4 m/s)

Two particles, with charges of 20nC and -20nC are placed at the points with coordinates (0,4.00 cm) and (0,-4.00 cm) as shown in the figure. A particle with charge 10 nC is located at the origin. A) Find the electric potential energy of the configuration of the three fixed charges. B) A fourth particle, with a mass of 2.00 x 10^-13 kg and a charge of 40.0 nC, is released from rest at the point (3.00 cm, 0). Find its speed after it has moved freely to a very large distance away.

D (points diagonally towards -Q due to attraction, and points diagonally away from +Q due to repulsion; the net force therefore points downwards)

What is the direction of the electrostatic force on charge +q? A) ← B) ↑ C) → D) ↓ E) None of the above F) need more information

D (electric field equation: E = kq/r^2) (you need a point where the electric fields will point in opposite directions, and will therefore cancel) (here it cannot be between the two points because both electric fields point towards the negative 2q charge; also A is incorrect because the q charge is smaller, so to the right of 2q it can never cancel with 2q because this charge is already larger, and then the distance of the smaller charge is always going to be greater than the 2q charge) (D is the correct answer because here q will be a shorter distance than the larger charge, so there will be a possible value where the electric fields can cancel)

Two point charges q and -2q lie on the x-axis as shown. Which region(s) on the x-axis include a point where the electric field due to the two point charges is zero? A) to the right of 2q B) between 2q and point P C) between point P and q D) to the left of q E) both A and C F) both B and D

perpendicular; distance V = 2πk*sigma *[(R^2+x^2)^1/2 - x]; Ex = 2πk*sigma*[1-x/[R^2+x^2]^1/2]

UNIFORMLY charged DISK (potential difference) 1) here dq = sigma*dA; and we integrate from 0 to R 2) the ring has a radius R and a surface charge density of sigma 3) P is along the ______ axis of the disk 4) P is on the central axis of the disk, symmetry indicates that all points in a give ring are the same ______ from P 5) calculations for the potential difference and electric field along x axis?

decreases (meaning work needed to move the charge is decreasing; more of a spontaneous process)

When a positive charge moves from point A to point B, the electric potential energy of the charge-field system _______.

Gaussian surface

a "closed" surface and the charge enclosed by such a surface

conductor

electrons are free to move in the material; here the electric field is ZERO inside the conductor and the charges must be residing on the surface