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Electric field as a function, find potential diff

  • Thread starter scholio
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  • #1
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Homework Statement


the electric field between two conducting surfaces is given by an expression:

E(x) = 9x^2 <i hat> where x is any location between the conductors. if one conductor is at x = 1 meter and the other is at x = 3 meter, what is the potential difference between them


Homework Equations



electric potential difference deltaV = V_B - V_A = - [<integral> E*dx] where x is distance

The Attempt at a Solution



i'm guessing i am supposed to do do the integral with the limits being from x = 1 to x = 3 so:
deltaV = [9x^3/3]x from x = 1 to x = 3, so ((9*27)/3)*3 - (9/3)*1 = 243 - 3 = +240 volts

i am not too sure about the correctness of my integration, but even still, was my approach correct?

thanks
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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deltaV = [9x^3/3]x
this is wrong. deltaV = [9x^3/3]
 
  • #3
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i thought my integration was wrong, i had it the way you have before, but doubted myself, well i redid the intrgral and did the calculation:

deltaV = [9x^3/3] from x = 1 to x = 3, so ((9*27)/3) - (9/3) = 81 - 3 = 78 volts

correct now?
thanks
 
  • #4
Redbelly98
Staff Emeritus
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Yep, looks good. :smile:
 

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