PHYS1220 Lecture 6: PHYS1220 Week 3.6

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aquamarinebadger817

10 Oct 2018

School

University of Newcastle

Department

Physics

Course

PHYS1220

Professor

Michael Gladys

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Document Summary

Electric potential due to a charged conducting sphere. For r > r0: v = k q/r. For r < r0: v = k q/r0. Single point charge: q at a distance r. An electric potential can be represented graphically by drawing equipotential lines (surfaces in 3d) The potential difference along an equipotential is zero and no work is required to move charges from point to point along the surface. An equipotential surface must be perpendicular to the electric field at any point. The rate of flow through a specified area of an object. Electric flux is the number of field lines going through the surface/volume. If the area is an irregular shape, flux is calculated using the surface integral. If the surface is formed in a closed shape, it is called a gaussian surface. If there is no charge inside a gaussian surface, the electric flux through it adds to zero.

Related Questions

1.Two uncharged, conducting spheres, A and B, are held at rest oninsulating stands and are in contact. A positively charged rod isbrought near sphere A. While the rod is in place, the two spheresare separated. How will the spheres be charged, if at all, afterthey are separated?
Sphere A Sphere B
(a) positive positive
(b) positive negative
(c) negative positive
(d) negative negative
(e) zero zero

2. A charged conductor is brought near an uncharged insulator.Which one of the following statements is true?
(a) Both objects will repel each other.
(b) Both objects will attract each other.
(c) Neither object exerts an electrical force on the other.
(d) The objects will repel each other only if the conductor has anegative charge.
(e) The objects will attract each other only if the conductor has apositive charge.

3. An electron is moved from point A, where the electric potentialis VA = â240 V, to point B, where the electric potential is VB =â360 V. What is the change in the electric
potential energy?
(a) 1.92Ã10â17 J (b) 2.04Ã10â17 J (c) 5.75Ã10â17 J (d) 8.98Ã10â17 J(e) 1.05Ã10â16 J

4. A -2q charge is located a distance d to the right of a +4qcharge, along the x
axis. At which point (or points) along the axis does the electricfield go to zero? To
solve this problem, consider an observer at point P, which is anarbitrary distance x from the +4q charge. Find the value (orvalues) for x (in terms of the separation distance d) where thetotal electric field vanishes.
(a) The electric field is never zero in the vicinity of thesecharges.
(b) The electric field is zero at x = -0.328d.
(c) The electric field is zero at x = +4.22d.
(d) The electric field is zero at both x = -2.26d and x =+2.26d.
(e) The electric field is zero at x = +3.41d.

5. point A is a distance L away from a point charge Q. Point B is adistance 4L away from Q. What is the ratio of the electrostaticpotential at B to that at A, i.e., what is VB/VA?
(a) 1/16
(b) 1/9
(c) 1/4
(d) 1/3
(e) can not be determined without specifying Q and L

6. Which one of the following statements best explains why it ispossible to define an electric potential
in a region of space that contains an electric field?
(a) Work must be done to bring two positive charges closertogether.
(b) Like charges repel one another and unlike charges attract oneanother.
(c) A positive charge will gain kinetic energy as it approaches anegative charge.
(d) The work required to bring two charges together is independentof the path taken.
(e) A negative charge will gain kinetic energy as it moves awayfrom another negative charge.

For a finite continuous charged object we can use integration tofind the electric potential at a point. As is the case for anyfinite charge distribution, the value of the electric potential canbe chosen to be zero at any reference point which is sufficientlyfar from the distribution. I.e. One can (and usually does) chooseV_(infinity) = 0 at r = infinity.

Consider a rod of length L extending from the origin along thepositive x axis, and a point P located at x = -D, which is adistance D from the origin. The task is to find the electricpotential (voltage) at the point P.

ChargedRod

The above Figure shows the rod. Highlighted is one small piece ofthe rod which is located at a position x and has length (dx). Note:In answers below, if you want to include dx, it is good practice toput (dx) in parentheses.

(a) Suppose that the rod is uniformly charged to a value Q. Writethe expression for the linear charge density, using the symbolsalready defined.
lambda = 1Click here to preview your answer.

(b) The small piece of the rod has length dx. In terms of symbolsL, x, (dx), D, Q and constants, what is the infinitesimal charge onthe rod? You may need to enclose dx in parentheses for it todisplay properly.
dq = 2Click here to preview your answer.

c) What is the distance from the point to the small piece of therod? Answer in terms of symbols x, L, D, (dx) and constants.
3Click here to preview your answer.

(d) What is the infinitesimal voltage produced at point P from thisdq? Answer in terms of symbols k, Q, x, L, D, (dx) andconstants.
dV = 4Click here to preview your answer.

(e) The final step would be to integrate this result. What are thelower 5Click here to preview your answer.
and upper 6Click here to preview your answer. limits on theintegration? Answer in terms of symbols x, L, D, (dx) andconstants.

Note that if the charge of the rod had not been uniform, we wouldneed to know lambda as a function of x and use that in place of(Q/L).

PHYS1220 Lecture 6: PHYS1220 Week 3.6

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