Four charges each Q in magnitude

[britt]

Homework Statement

Four charges, each q in magnitude, are arranged in a square, two positive and two negative, in opposite corners. The square is L length on each side, and the charges are all in the same plane as shown in the diagram. Show work and units for full credit.

q+ -----B----- q-
| |
| | L
| |
q- ------------ q+
L
**vertical dotted line makes square but spacing corrects itself

A. Determine the magnitude and direction of the total electric field at point B (half way between the top two charges) due to all four charges as a function of q and L.

B. If placed at point B, which would accelerate faster, object 1 with a charge 2Q and mass M or object 2 with a charge Q and mass M/2? Why?

C. Draw the electric field lines in the plane of the square indicating the direction and general shape of the electric field generated by this charge arrangement. Hint: think about the electric field generated by two opposite charges.

D. Is the electric field zero at any point(s)? If so, where? Also, indicate any equilibrium point(s) and whether they are stable, unstable or neutral.

The Attempt at a Solution

A. E=(36/5)(q/L^2)

B. Both are equal a = 2qE/m = qE/(m/2)

D. 0 at top of square going in towards B from + and towards B for -

 

[BOYLANATOR]

What equation are you using for electric field strength?

 

[britt]

E=k(q/R^2)

and adding

 

[britt]

I got that from the bottom charges R=(sqrt(5)/2)L

and the other two cancel out because they go in towards B

E=k(q/(5/4)L^2)

 

[lightgrav]

the E-field contributions by the top charges are both rightward:
away from +Q , and toward -Q (toward being "negative away"),
but some of this is canceled by leftward E-field by the bottom charges.
What fraction of the E-field contribution by the bottom left -Q is matched by the +Q on bottom right?

 

[britt]

The bottom left -Q and +Q would result in a leftward towards upper +Q but I thought the bottom ones would cancel out completely?

 

[lightgrav]

away from the positive and toward the negative; only their vertical components cancel ... their horizontals double.

 

[britt]

so their force added would be 8kq/5L^2 .. ?

 

[lightgrav]

each E-field contribution is 4/5 kQ/L² , but that is along the diagonal. You need to find the horizontal component.

 

[britt]

horizontal component: 2(4/5)(kQ/L^2)cos(180)+2(4/5)(kQ/L^2)cos(0)

 

[lightgrav]

Look at the triangle that you used to find the diagonal distance. The E-field contribution is along that diagonal.
The negative's contribution is mostly down, but a little bit leftward ... left is L/2 compared to the diagonal L√5 /2 ... that is, the diagonal contribution is √5 times as long as the horizontal part that survives.

 

[britt]

So would the x component be:

Lcos(180)+Lcos(0)?
Ok (duh) it's running from b to the top charge(l/2)
Thanks for your help I never understand the x and y components

 

[lightgrav]

/
/|
/ |
/_|
the Ex/E ratio is the same as the Lx/R ratio ... each = sin theta (the skinny angle)
so Ex = E Lx/R , leftward (negative)
The E-field contributions from the TOP charges are at 0° (both are rightward)

 

[lightgrav]

|\
| \ E from bottom right +Q , upward
| \
|__\
Ex

sin theta = Ex/E , but also sin theta = ½L /R ... so Ex = ½L /R * E

 

[britt]

Okay so add ((L/2)/R)xE and EL/R then multiply by cos(180) + (8kq/5L^2) ?

 

[britt]

Or is the component 1/2L/R *E ?

 

[lightgrav]

yes, pointing leftward, if you put the correct expression in for E. and R.
then you add these (subtract, actually) to the top Q's E-fields ; don't forget those, because they're much bigger!

 

[britt]

Yes to the first post

Okay so add ((L/2)/R)xE and EL/R then multiply by cos(180) + (8kq/5L^2) ?

Correction at end:: + (8kq/5L^2)cos(0)


Or two the Ex=(1/2)L/R * E ?

 

[lightgrav]

where did the 8 in the 8/5 come from?

don't confuse yourself by writing "cos(180)" when you mean leftward.
the triangle base is ½ ; the diagonal is 1.118 , so the base is .5/1.118 times as big.
the same ratio applies to the red arrows (E-field contributions)
http://www.science.marshall.edu/foltzc/addEsq.gif