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Why is this true?
\[\vec E \cdot d \vec l=E \hat i \cdot (dx \hat i +dy \hat j +dz \hat k)=E dx\]
I understand the first part
\[\vec E \cdot d \vec l=E \hat i \cdot (dx \hat i +dy \hat j +dz \hat k)\] but how does that equal Edx?
 one year ago
 one year ago
Why is this true? \[\vec E \cdot d \vec l=E \hat i \cdot (dx \hat i +dy \hat j +dz \hat k)=E dx\] I understand the first part \[\vec E \cdot d \vec l=E \hat i \cdot (dx \hat i +dy \hat j +dz \hat k)\] but how does that equal Edx?
 one year ago
 one year ago

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JenniferSmart1Best ResponseYou've already chosen the best response.1
@zepdrix @hartnn
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
i and j are perpendicular i and k are perpendicular
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
i.i = i^2 = 1 i.j =0 i.k=0
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
I'm sorry...but how did you determine that they were perpendicular?
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
i , j and k are standard unit vectors. i is along x direction j is along y direction k is along z direction and axes are perpendicular to each other, right ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
dw:1361080799293:dw any more doubts ?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
ahhhh.....nope...no more doubts =D
 one year ago
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