2. The electric field is given by the equation E- cy?i+ 2e xyî where c 1000 V/m3. (a) How much work does the electrical field do on a charge of-2 μC taken from point (0,0) along thex axis and then parallel to the y-axis to (1m, 2 m)? (b) How much electrical work is done when the cha moved along a straight line from (0, 0) to (1 m, 2 m)? (c) What is the equation for the electri of this field assuming the origin has a potential of zero? (d) What is the electric potential at point to (1 m, 0) tial (1 m, 2 Solution: Given: q-_2.00 x 10-6 C Drawing c 1000 V/m3 Equations: For path AB: y = 0, dy = 0, and x goes from 0 to 1 m, so WAR-0. For path BC x=lm, dx = 0, and y goes from 0 to 2 m, so wBC-xqcJ2ydy=xqc[y21 8.00x10-3 J For total path WAse Wa+ Wne 8.0 x10" J (0,0) 2m (a) 2m (b) For path AC: y 2x, dy-2 dx, and x goes from 0 to l m, so 4x3 8x311m 38.00x10-3 J (c) V =-JE-dǐ--(fExdx +fEydy)--cU32.dx +fzxydy constant with dx-0 while y goes from 0 to y. For the first part of this path, V-0, and for the second part This must be true along any path, so choose the path y 0, dy- 0, while x goes from 0 to x; then hold x (d) V(l m, 2 m) -4000v
2. The electric field is given by the equation E- cy?i+ 2e xyî where c 1000 V/m3. (a) How much work does the electrical field do on a charge of-2 μC taken from point (0,0) along thex axis and then parallel to the y-axis to (1m, 2 m)? (b) How much electrical work is done when the cha moved along a straight line from (0, 0) to (1 m, 2 m)? (c) What is the equation for the electri of this field assuming the origin has a potential of zero? (d) What is the electric potential at point to (1 m, 0) tial (1 m, 2 Solution: Given: q-_2.00 x 10-6 C Drawing c 1000 V/m3 Equations: For path AB: y = 0, dy = 0, and x goes from 0 to 1 m, so WAR-0. For path BC x=lm, dx = 0, and y goes from 0 to 2 m, so wBC-xqcJ2ydy=xqc[y21 8.00x10-3 J For total path WAse Wa+ Wne 8.0 x10" J (0,0) 2m (a) 2m (b) For path AC: y 2x, dy-2 dx, and x goes from 0 to l m, so 4x3 8x311m 38.00x10-3 J (c) V =-JE-dǐ--(fExdx +fEydy)--cU32.dx +fzxydy constant with dx-0 while y goes from 0 to y. For the first part of this path, V-0, and for the second part This must be true along any path, so choose the path y 0, dy- 0, while x goes from 0 to x; then hold x (d) V(l m, 2 m) -4000v
College Physics
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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Transcribed Image Text:2. The electric field is given by the equation E- cy?i+ 2e xyî where c 1000 V/m3.
(a) How much
work does the electrical field do on a charge of-2 μC taken from point (0,0) along thex axis
and then parallel to the y-axis to (1m, 2 m)? (b) How much electrical work is done when the cha
moved along a straight line from (0, 0) to (1 m, 2 m)? (c) What is the equation for the electri
of this field assuming the origin has a potential of zero? (d) What is the electric potential at point
to (1 m, 0)
tial
(1 m, 2
Solution:
Given: q-_2.00 x 10-6 C
Drawing
c 1000 V/m3
Equations:
For path AB: y = 0, dy = 0, and x goes from 0 to 1 m, so WAR-0.
For path BC x=lm, dx = 0, and y goes from 0 to 2 m, so wBC-xqcJ2ydy=xqc[y21
8.00x10-3 J
For total path WAse Wa+ Wne 8.0 x10" J
(0,0)
2m
(a)
2m
(b)
For path AC: y
2x, dy-2 dx, and x goes from 0 to l m, so
4x3 8x311m
38.00x10-3 J
(c) V =-JE-dǐ--(fExdx +fEydy)--cU32.dx +fzxydy
constant with dx-0 while y goes from 0 to y. For the first part of this path, V-0, and for the second part
This must be true along any path, so choose the path y
0, dy- 0, while x goes from 0 to x; then hold x
(d) V(l m, 2 m) -4000v
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