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ketz

  • 3 years ago

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.

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  1. ketz
    • 3 years ago
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  2. ketz
    • 3 years ago
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    Urgent answer needed!! (I've exams tomorrow)

  3. imron07
    • 3 years ago
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    Do you understand vector notation?

  4. pkjha3105
    • 3 years ago
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    it shows the direction of vector quantity...

  5. ketz
    • 3 years ago
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    What about the magnitude?

  6. NoelGreco
    • 3 years ago
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    \[E=k \frac{ Q }{ r ^{2} }\] Calculate the magnitude of the electric field at point B

  7. Algebraic!
    • 3 years ago
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    |dw:1348614053138:dw|

  8. Algebraic!
    • 3 years ago
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    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

  9. Algebraic!
    • 3 years ago
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    any questions about that?

  10. woleraymond
    • 3 years ago
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    @Algebraic! PLEASE MAKE IT MORE CLEARER

  11. Algebraic!
    • 3 years ago
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    what part?

  12. Algebraic!
    • 3 years ago
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    finding magnitudes of E?

  13. Algebraic!
    • 3 years ago
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    E1: (9E9*7E-6)/.005m etc

  14. woleraymond
    • 3 years ago
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    YEA..I WAS THINK THE 2 CHARGES WILL ADD UP ALGEBRAICALLY AND THE RESULTANT WILL HAVE EFFECT ON THE TEST CHARGE PROTON

  15. Algebraic!
    • 3 years ago
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    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!!_

  16. woleraymond
    • 3 years ago
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    PLS HOW DID THE ANGLE BECOME 45 DEGREES

  17. Algebraic!
    • 3 years ago
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    well...

  18. Algebraic!
    • 3 years ago
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    |dw:1348615004630:dw|

  19. Algebraic!
    • 3 years ago
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    that's better

  20. Algebraic!
    • 3 years ago
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    does that clear it up?

  21. woleraymond
    • 3 years ago
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    yep it does

  22. Algebraic!
    • 3 years ago
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    so you got E2 (mag. and direction/components) ?

  23. woleraymond
    • 3 years ago
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    @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

  24. Algebraic!
    • 3 years ago
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    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)

  25. woleraymond
    • 3 years ago
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    i really learnt alot from you @Algebraic!

  26. Algebraic!
    • 3 years ago
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    haha! good! I hope so!

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