Finding electric field intensity

[Prathamesh]

3 point charges q1 q2 & q3 are placed in vacuum as shown.
Find electric field intensity at B,if q1=2C,q2=-4C & q3=-2C

Homework Equations

The Attempt at a Solution

 

[phoenix95]

First find the electric field intensity at the point due individual charges, then find the resultant.

Note:
Electric field intensity due to a charge q at a distance r in the direction of r is given by
E=(1/4πε₀)(q/r³)r
ε₀=8.854×10^-12 Farads/m

 

[Prathamesh]

I have tried it up to this...
Is it right ?
If yes , what to do next and how?
Plz help

 

[Prathamesh]

First find the electric field intensity at the point due individual charges, then find the resultant.

Note:
Electric field intensity due to a charge q at a distance r in the direction of r is given by
E=(1/4πε₀)(q/r³)r
ε₀=8.854×10^-12 Farads/m

Thank you!
But while finding resultant vector , can I first find resultant of intensity (E') due to q2 & q3 and then simply subtract intensity due to q1 (E1) from E'(though E' is not exactly opposite to E1)?

 

[phoenix95]

Hey, are you sure the units of charge are in C, not nC or anything?
If it is C, then the Es you calculated are wrong... You have write it in terms powers of ten, otherwise use the correct units. (Keep in mind the electrostatic force due to one coulomb of charge is extremely large)

But while finding resultant vector , can I first find resultant of intensity (E') due to q2 & q3 and then simply subtract intensity due to q1 (E1) from E'(though E' is not exactly opposite to E1)?

Keeping in mind the directions of Es, that is a correct way to do it...

 

[Prathamesh]

Hey, are you sure the units of charge are in C, not nC or anything?
If it is C, then the Es you calculated are wrong... You have write it in terms powers of ten, otherwise use the correct units. (Keep in mind the electrostatic force due to one coulomb of charge is extremely large)Keeping in mind the directions of Es, that is a correct way to do it...

Yeah... sorry , Charges are in nC.

so , finding resultant is like this? In solutions it is given that even there is angle θ(θ≠180) between 2 vectors , simply subtract them...
I didnt understand that...

 

[phoenix95]

so , finding resultant is like this?

Yes.

In solutions it is given that even there is angle θ(θ≠180) between 2 vectors , simply subtract them...
I didnt understand that...

First see if you get the correct resultant . I didn't catch the 'angle' thing... If you get correct answer then will you please post what exactly the solution says? I'll try my best to explain...

 

[Prathamesh]

Yes.First see if you get the correct resultant . I didn't catch the 'angle' thing... If you get correct answer then will you please post what exactly the solution says? I'll try my best to explain...

Solution says after finding E' (which comes 18*√5=40.248N/C) to get final result subtract E1 (9N/C) from it. Thus final result becomes 31.248 N/C.
But I think,to get final result vector sum of E' and E1 should be done which gives final result =√1620+81+2*18√5*9*cosθ=√ 1701+362.232cosθ

 

[phoenix95]

Solution says after finding E' (which comes 18*√5=40.248N/C) to get final result subtract E1 (9N/C) from it. Thus final result becomes 31.248 N/C.
But I think,to get final result vector sum of E' and E1 should be done which gives final result =√1620+81+2*18√5*9*cosθ=√ 1701+362.232cosθ

Correct. In order do what the solution says, the vectors E' and E₁ have to be exactly opposite. You are right.

 

[Prathamesh]

since cosθ comes -0.94≈-1 , it is approximately same as subtracting E1 from E' ...
Thank you...