Determine work done by electric field

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SUMMARY

The discussion centers on calculating the electric potential difference \( V_{MN} \) between points M(2,6,-1) and N(-3,-3,2) in the electric field \( \vec E = 6x^2\hat i + 6y\hat j + 4\hat z \) v/m. The user initially attempted to find \( V_{MN} \) using a normalized displacement vector, leading to an incorrect result of -103.97 V, while the textbook solution is -139.0 V. The correct approach involves using the displacement vector \( d\vec \ell = \hat i dx + \hat j dy + \hat k dz \) without normalization to accurately compute the integral.

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Homework Statement


Given electric field ##\vec E = 6x^2\hat i +6y\hat j+4\hat z## v/m
Find ## V_{MN}## if both M and N separate by M(2,6,-1) and N(-3,-3,2)

Homework Equations



The Attempt at a Solution


here i find ##V_{MN}## by
unit vector from N to M is ##\frac 1 {\sqrt (115)}(5\hat i +9\hat j -3\hat k)##
so ##V_{MN} = -\int \vec E \cdot \vec {dl} = \int \vec E \cdot \vec {\frac 1 {\sqrt 115}(5dx\hat i +9dy\hat j -3dz\hat k)}##
and then i find intregral with respect to x from -3 to 2
respect to y from -3 to 6
respect to z from 2 to -1
and my answer is -103.97 v but textbook solution is -139.0 v what i do wrong
 
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You should not normalise ##d\vec \ell##. You need to use the appropriate expression for the displacement vector, i.e., ##d\vec \ell = \hat i dx + \hat j dy + \hat k dz##.
 
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