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 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
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?

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JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1@zepdrix @hartnn

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

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1I'm sorry...but how did you determine that they were perpendicular?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2i , 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 ?

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1ahhhh.....nope...no more doubts =D
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