ketz Group Title 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. one year ago one year ago

1. ketz Group Title

2. ketz Group Title

Urgent answer needed!! (I've exams tomorrow)

3. imron07 Group Title

Do you understand vector notation?

4. pkjha3105 Group Title

it shows the direction of vector quantity...

5. ketz Group Title

What about the magnitude?

6. NoelGreco Group Title

$E=k \frac{ Q }{ r ^{2} }$ Calculate the magnitude of the electric field at point B

7. Algebraic! Group Title

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8. Algebraic! Group Title

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! Group Title

any questions about that?

10. woleraymond Group Title

@Algebraic! PLEASE MAKE IT MORE CLEARER

11. Algebraic! Group Title

what part?

12. Algebraic! Group Title

finding magnitudes of E?

13. Algebraic! Group Title

E1: (9E9*7E-6)/.005m etc

14. woleraymond Group Title

YEA..I WAS THINK THE 2 CHARGES WILL ADD UP ALGEBRAICALLY AND THE RESULTANT WILL HAVE EFFECT ON THE TEST CHARGE PROTON

15. Algebraic! Group Title

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 Group Title

PLS HOW DID THE ANGLE BECOME 45 DEGREES

17. Algebraic! Group Title

well...

18. Algebraic! Group Title

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19. Algebraic! Group Title

that's better

20. Algebraic! Group Title

does that clear it up?

21. woleraymond Group Title

yep it does

22. Algebraic! Group Title

so you got E2 (mag. and direction/components) ?

23. woleraymond Group Title

@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! Group Title

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 Group Title

i really learnt alot from you @Algebraic!

26. Algebraic! Group Title

haha! good! I hope so!