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anonymous
 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.
anonymous
 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Urgent answer needed!! (I've exams tomorrow)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you understand vector notation?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it shows the direction of vector quantity...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What about the magnitude?

NoelGreco
 3 years ago
Best ResponseYou've already chosen the best response.0\[E=k \frac{ Q }{ r ^{2} }\] Calculate the magnitude of the electric field at point B

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348614053138:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0E1 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
 3 years ago
Best ResponseYou've already chosen the best response.0any questions about that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! PLEASE MAKE IT MORE CLEARER

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0finding magnitudes of E?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0E1: (9E9*7E6)/.005m etc

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0YEA..I WAS THINK THE 2 CHARGES WILL ADD UP ALGEBRAICALLY AND THE RESULTANT WILL HAVE EFFECT ON THE TEST CHARGE PROTON

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0direction of E1 : angle is 45 so E1x = ((9E9*7E6)/.005m) *cos(45) E1y =((9E9*7E6)/.005m) *sin(45) (edited: now with more parentheses!!_

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0PLS HOW DID THE ANGLE BECOME 45 DEGREES

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348615004630:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0does that clear it up?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you got E2 (mag. and direction/components) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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
 3 years ago
Best ResponseYou've already chosen the best response.0yeah 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
 3 years ago
Best ResponseYou've already chosen the best response.0i really learnt alot from you @Algebraic!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha! good! I hope so!
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