ketz
  • ketz
Use Coulomb’s Law to determine the magnitude and direction of the electric field EB at point B in the diagram due to the two positive charges (Q = 7.0 μC). If a proton is located at point B, calculate the magnitude and direction of the force F acting on the proton.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ketz
  • ketz
1 Attachment
ketz
  • ketz
Urgent answer needed!! (I've exams tomorrow)
anonymous
  • anonymous
Do you understand vector notation?

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anonymous
  • anonymous
it shows the direction of vector quantity...
ketz
  • ketz
What about the magnitude?
NoelGreco
  • NoelGreco
\[E=k \frac{ Q }{ r ^{2} }\] Calculate the magnitude of the electric field at point B
anonymous
  • anonymous
|dw:1348614053138:dw|
anonymous
  • anonymous
E1 has the same direction as r1 E2 has the same direction as r2 find the magnitudes of each field in terms of Q find the 'x' and 'y' components of each field add the 'y' components; add the x components : those sums are the components of the net field
anonymous
  • anonymous
any questions about that?
anonymous
  • anonymous
@Algebraic! PLEASE MAKE IT MORE CLEARER
anonymous
  • anonymous
what part?
anonymous
  • anonymous
finding magnitudes of E?
anonymous
  • anonymous
E1: (9E9*7E-6)/.005m etc
anonymous
  • anonymous
YEA..I WAS THINK THE 2 CHARGES WILL ADD UP ALGEBRAICALLY AND THE RESULTANT WILL HAVE EFFECT ON THE TEST CHARGE PROTON
anonymous
  • anonymous
direction of E1 : angle is 45 so E1x = ((9E9*7E-6)/.005m) *cos(45) E1y =((9E9*7E-6)/.005m) *sin(45) (edited: now with more parentheses!!_
anonymous
  • anonymous
PLS HOW DID THE ANGLE BECOME 45 DEGREES
anonymous
  • anonymous
well...
anonymous
  • anonymous
|dw:1348615004630:dw|
anonymous
  • anonymous
that's better
anonymous
  • anonymous
does that clear it up?
anonymous
  • anonymous
yep it does
anonymous
  • anonymous
so you got E2 (mag. and direction/components) ?
anonymous
  • anonymous
@Algebraic! yep its clear on how to get E2 but the question also want the force F acting on the proton also after getting the x and y components of E1 and E 2 how will it be added
anonymous
  • anonymous
yeah once you find the net field at a point, the force on a particle at that point is just qE, where q is the charge of the particle (the charge of a proton in this case: +e) "also after getting the x and y components of E1 and E 2 how will it be added" just add them! add the x comp of E1 and the x comp of E2, that's the x comp of the net field... repeat for y comp to find the magnitude of the net field, use sqrt( (x comp)^2 + (y comp)^2 ) to find the direction of the net field, use arctan( x comp / y comp)
anonymous
  • anonymous
i really learnt alot from you @Algebraic!
anonymous
  • anonymous
haha! good! I hope so!

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