Pledged Problem 4 Solutions: Electric Potential and Electric Field in Physics 102, Study notes of Physics

Solutions to pledged problem 4 in physics 102, which involves calculating the electric potential and electric field for various charge distributions. The problem includes a point charge and an infinite non-conducting planar sheet of charge. The solutions are based on the concepts of potential difference, gauss' law, and symmetry.

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Pre 2010

Uploaded on 08/16/2009

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Physics 102 – Pledged Problem 4
Time allowed: 2 hours at a single sitting
DUE 4PM MONDAY, February 11, 2008, in the boxes marked Phys 101-102 in the physics lounge.
You may use your own textbook, your notes, and a non-programmed calculator. You may also consult
the on-line solutions to the corresponding suggested problems. You should consult no other help. Show
how you arrived at your answer; the correct answer by itself may not be sufficient.
Further instructions:
(a) Write legibly on one side of 8.5” x 11” white or lightly tinted paper.
(b) Staple all sheets together, including this one, in the upper left corner and make one vertical fold.
(c) On the outside, staple side up. Print your name in capital letters, your LAST NAME first
followed by your FIRST NAME.
(d) Below your name, print the phrase “Pledged Problem 4”, followed by the due date.
(e) Also indicate start time and end time.
(f) Write and sign the pledge, with the understanding you may consult the materials noted above.
____________________________________________________________________________________
1. (4 pts) A charge +4q is located at the origin and a charge –q is on the x axis at x=a.
(a) Write an expression for the potential on the x axis for x > a;
(b) Find a point in this region where V = 0;
(c) Use the result of (a) to find the electric field
󰇍
󰇍
󰇍
on the x axis for x > a; and
(d) Find a point where E = 0.
2. (6 pts) Consider an infinite non-conducting planar sheet of charge of uniform charge density ρ
that lies in the XZ plane and has thickness b (see figure).
(a) Use Gauss’ Law to calculate the electric field (vector) as a function of the y coordinate, i.e.
the coordinate perpendicular to the plane of the charge sheet. (The origin is taken to be at the
center of the sheet).
(b) Calculate the resulting potential as a function of y. Set the potential at y = 0 to be zero.
Sketch this potential.
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Physics 102 – Pledged Problem 4

Time allowed: 2 hours at a single sitting

DUE 4PM MONDAY, February 11, 2008, in the boxes marked Phys 101-102 in the physics lounge. You may use your own textbook, your notes, and a non-programmed calculator. You may also consult the on-line solutions to the corresponding suggested problems. You should consult no other help. Show how you arrived at your answer; the correct answer by itself may not be sufficient.

Further instructions:

(a) Write legibly on one side of 8.5” x 11” white or lightly tinted paper. (b) Staple all sheets together, including this one, in the upper left corner and make one vertical fold. (c) On the outside, staple side up. Print your name in capital letters, your LAST NAME first followed by your FIRST NAME. (d) Below your name, print the phrase “Pledged Problem 4”, followed by the due date. (e) Also indicate start time and end time. (f) Write and sign the pledge, with the understanding you may consult the materials noted above.


  1. (4 pts) A charge +4q is located at the origin and a charge –q is on the x axis at x=a.

(a) Write an expression for the potential on the x axis for x > a;

(b) Find a point in this region where V = 0;

(c) Use the result of (a) to find the electric field ܧሬሬሬԦon the x axis for x > a; and

(d) Find a point where E = 0.

  1. (6 pts) Consider an infinite non-conducting planar sheet of charge of uniform charge density ρ

that lies in the XZ plane and has thickness b (see figure).

(a) Use Gauss’ Law to calculate the electric field (vector) as a function of the y coordinate, i.e.

the coordinate perpendicular to the plane of the charge sheet. (The origin is taken to be at the

center of the sheet).

(b) Calculate the resulting potential as a function of y. Set the potential at y = 0 to be zero.

Sketch this potential.

Solution – Pledged Problems

ଵ ସగ

  1. (a) This problem requires applying the concept that the total potential at a point is the sum of the potentials associated with each charge element sourcing the potential -

ఌ (^) బ^ ∑^

௤೔ ௥೔

௡ ௜ୀଵ. So, for the charge distribution in the problem,^ ܸ ௉ for^ ܽ൐ ݔ^ is

ܸ ௉ ሺݔሻ ൌ

ሺbሻ ܸ (^) ௉ ሻݔሺ ସ is zero when ቀ௫ െ ଵ ሺ௫ି௔ሻ ቁ ൌ 0^

ܧ

, or when ൌ ݔ. (Plugging this value for x back into the equation confirms this is the right answer).

(c) The electric field on the x -axis, ܧ௫, is derivable from the potential through the relationship ௫ ൌ െ^

ௗ௏ ௗ௫. Taking the derivative of^ ܸ ௉^ ሻݔሺ^ with respect to^ x^ yields – ൭

ܧ௫ ൌ െ ݀݀ ݔ ቇ൱ ൌ െ

݀݀ ݔ ቇ

ݔ^2

ቇ ସ (d) ܧ௫ ൌ 0 when ቀ௫ (^) మെ ଵ ሺ௫ି௔ ሻమቁ ൌ 0, 4ሺ ݔെܽ ሻ ଶ^ ݔ െ ଶ^ ݔ3 ൌଶ^ ܽ4 ൅ ݔܽ8 െ ଶ^ ൌ 0

ሺ ݔെ 2ܽ ሻሺ3 ݔെ 2 ܽሻ ൌ 0

ܽ0 ൌ ܽ ௫

The location where ܧ௫ for ൐ ݔ is, therefore, ൌ ݔ. (Again, plugging this value for x into the equation for ܧ confirms this is the right answer).

  1. Symmetry of the charge distribution dictates the type of surface to be used in applying Gauss’ Law. As we’ve discussed in class and in the book, a planar uniform charge distribution will produce electric fields that are perpendicular to the surface and of equal magnitude a given distance from the surface. In all regions, the direction of E will be away from the x-axis; so for y > 0, the direction is ൅ଔ̂, and for y<0, will be െଔ̂. This symmetry is true inside and outside the planar sheet, so we will choose a structure whose sides are parallel to the electric fields, and whose top and bottom are equidistant from the y- axis. Φ 1 is flux through a representative surface inside the sheet; Φ 2 outside the sheet.

r ܾെ ൏ ݕ ൗ 2 to be the same as ܾ൐ ݕ ൗ 2. So, a sketch

of y v. V(y) will look like

Symmetry requires the potential fo