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.

- ketz

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- ketz

##### 1 Attachment

- ketz

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

- anonymous

Do you understand vector notation?

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## More answers

- anonymous

it shows the direction of vector quantity...

- ketz

What about the magnitude?

- NoelGreco

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

- anonymous

|dw:1348614053138:dw|

- 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

any questions about that?

- anonymous

@Algebraic! PLEASE MAKE IT MORE CLEARER

- anonymous

what part?

- anonymous

finding magnitudes of E?

- anonymous

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

- anonymous

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

- 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

PLS HOW DID THE ANGLE BECOME 45 DEGREES

- anonymous

well...

- anonymous

|dw:1348615004630:dw|

- anonymous

that's better

- anonymous

does that clear it up?

- anonymous

yep it does

- anonymous

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

- 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

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

i really learnt alot from you @Algebraic!

- anonymous

haha! good! I hope so!

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