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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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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.
(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.
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.
ଵ ସగ
ఌ (^) బ^ ∑^
ୀଵ. So, for the charge distribution in the problem,^ ܸ for^ ܽ ݔ^ is
ܸ ሺݔሻ ൌ
ሺbሻ ܸ (^) ሻݔሺ ସ is zero when ቀ௫ െ ଵ ሺ௫ିሻ 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
ቆ
ܧ௫ ൌ െ ݀݀ ݔ ቇ൱ ൌ െ
ቆ
݀݀ ݔ ቇ
ቆ
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).
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