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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
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

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ketz Group TitleBest ResponseYou've already chosen the best response.0
Urgent answer needed!! (I've exams tomorrow)
 one year ago

imron07 Group TitleBest ResponseYou've already chosen the best response.0
Do you understand vector notation?
 one year ago

pkjha3105 Group TitleBest ResponseYou've already chosen the best response.0
it shows the direction of vector quantity...
 one year ago

ketz Group TitleBest ResponseYou've already chosen the best response.0
What about the magnitude?
 one year ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
dw:1348614053138:dw
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
any questions about that?
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@Algebraic! PLEASE MAKE IT MORE CLEARER
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
what part?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
finding magnitudes of E?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
E1: (9E9*7E6)/.005m etc
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
YEA..I WAS THINK THE 2 CHARGES WILL ADD UP ALGEBRAICALLY AND THE RESULTANT WILL HAVE EFFECT ON THE TEST CHARGE PROTON
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
direction of E1 : angle is 45 so E1x = ((9E9*7E6)/.005m) *cos(45) E1y =((9E9*7E6)/.005m) *sin(45) (edited: now with more parentheses!!_
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
PLS HOW DID THE ANGLE BECOME 45 DEGREES
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
well...
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
dw:1348615004630:dw
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
that's better
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
does that clear it up?
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
yep it does
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
so you got E2 (mag. and direction/components) ?
 one year ago

woleraymond Group TitleBest 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
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
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)
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
i really learnt alot from you @Algebraic!
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
haha! good! I hope so!
 one year ago
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