2A Pod f lemth Lis lying on x-axis with let end at the origin and has a non-unigomly lincak chaege donsity of 2= 5x Frd the clecteic potontjal V at A (-4,0) dV - Kida R=X+4. Anti dV= , dg= 2de dV= K2 dx dV- h 5x dx x+4 メ+4 V= 5K U=X+4 du-ch ヤUS0:EU-4 →ラんJ U-4 o do + ん 1 5K [S1 du - 4S J du] 5K [(x) -4 In1x+4l + 5k (L+4-4in[ L+4D) - (4 -4 In 4D] 5K|L] → V= 4.5 x 10"L volts -> 10
2A Pod f lemth Lis lying on x-axis with let end at the origin and has a non-unigomly lincak chaege donsity of 2= 5x Frd the clecteic potontjal V at A (-4,0) dV - Kida R=X+4. Anti dV= , dg= 2de dV= K2 dx dV- h 5x dx x+4 メ+4 V= 5K U=X+4 du-ch ヤUS0:EU-4 →ラんJ U-4 o do + ん 1 5K [S1 du - 4S J du] 5K [(x) -4 In1x+4l + 5k (L+4-4in[ L+4D) - (4 -4 In 4D] 5K|L] → V= 4.5 x 10"L volts -> 10
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Question
![potential Vat A (-4,0)
2)d20d lemth Lis lying on x-axis with let end at the origin and
Pod
bas a non-uniyoemly lincae chacge donsity f 2= 5x
7nd the electeic'poontial Vat A (-4,0)
dV-K.da R=X+4.
ORigin and
lineaR chaRge
densi
23D5X
Anti dV= , dg= 2 dx
dV- K2 dx
dV3 h 5x dx
4>
4>
x+4
メ+4
V= 5KS"
5K LSi du - 45Jdu 5k [ (x)-4 In/x+41
5k/(L+4-41n/L+41)-(4-4 In141)]
5k[L]→V=4,5 ×10"L volts
U-4
C.
U=X+H du-ch
ヤUSU0 x=U-4
カ5%J 5do
10
%=D4.5x1O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07b04562-0d10-441a-9a5d-597bbf786c4c%2F87706c8b-991d-4992-b3dd-f62121312df4%2F2utcopt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:potential Vat A (-4,0)
2)d20d lemth Lis lying on x-axis with let end at the origin and
Pod
bas a non-uniyoemly lincae chacge donsity f 2= 5x
7nd the electeic'poontial Vat A (-4,0)
dV-K.da R=X+4.
ORigin and
lineaR chaRge
densi
23D5X
Anti dV= , dg= 2 dx
dV- K2 dx
dV3 h 5x dx
4>
4>
x+4
メ+4
V= 5KS"
5K LSi du - 45Jdu 5k [ (x)-4 In/x+41
5k/(L+4-41n/L+41)-(4-4 In141)]
5k[L]→V=4,5 ×10"L volts
U-4
C.
U=X+H du-ch
ヤUSU0 x=U-4
カ5%J 5do
10
%=D4.5x1O
Expert Solution
Step 1
The given rod of length L lies along the x axis, with one of its ends at the origin.
The point A is situated at a point (-4,0), which lies along the negative x axis, as shown in the figure
Step by step
Solved in 5 steps with 1 images
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