Pls help .. Electric field question?

Question

Find the charge Q that should be placed at the centre of the square of side 1.00E+1 cm, at the corners of which four identical charges +q = 3 C are placed so that the whole system is in equilibrium

the figure of the question : http://www.upload-picture.com/viewimage.php?file=/images/nws42291.gif

any one could help me pls ???

Answer 1

Looking at a single charge q, is repelled by the three others. To make it in equilibrium it needs to be attracted by a charge Q such that the force of attraction of Q counteracts the force of repulsion of the 3 charges.

The force of repulsion between 2 charges is

K = kc qq/r^2

let r be the side length of the square
let X be the horizontal unit vector increasing toward the right
let Y be the vertical unit vector increasing toward the top
let F be the force felt by the top-right most charge q

The force from the charge to the left is K X
The force from the charge below is K Y
The force from the charge in diagonal is

kc qq/(sqrt2.r)^2 sin45 X + kc qq/(sqrt2.r)^2 cos45 Y =
K/2 sin45 X + K/2 cos45 Y =
K/2Sqrt2 X + K/2Sqrt2 Y

So total force is

F = (K + K/(2sqrt2)).X + (K + K/(2sqrt2)).Y
F = (K.(1+1/(2sqrt2))) . (X + Y)

Or in polar notation
F = sqrt(2).K.(1+1/(2sqrt2)) with an angle of 45 degree
F = K.(sqrt(2) + 1/2) with an angle of 45 degree

To exactly compensated, a charge Q is positioned at angle -45 degree and distance R = sqrt(2)/2 . r. This will create a force G of magnitude

G = kc qQ/(r.sqrt(2)/2)^2 = 2.kc.qQ/r^2 with an angle of -45 degree

therefore we have

2.kc.qQ/r^2 = -K.(sqrt(2) + 1/2)
2.kc.qQ/r^2 = -kc qq/r^2 . (sqrt(2) + 1/2)
2.Q = -q . (sqrt(2) + 1/2)
Q = -q . (sqrt(2) + 1/2)/2 =
Q = -0.957 q = -2.87 C

So putting a charge 0f -2.87 C in the middle of the square would equilibrate everything.

Answer 2

Let these be named as q1= q2= q3= q4 = 3C

Just suppose q1= q3= 0 are zero or absent from the same diagonal.

The remaining charges (q2=q4=3c) on same diagonal
can be balanced at equilibrium by placing at center ( - 6c) thus the resulting QUADRUPOLE will be under equilibrium.

Now it you bring in charge q1 near this quadrupole then balance will be effected so we can bring in both q1 and q3 from respective diagonal paths and this balance will remain unaffected due to 2 cross quadrupoles - whatever be the dimensions of the square.

As long as 4 charges are same =q then the equilibrium of system through center will always be achieved by placing opposite double magnitude charge there.

Nevertheless, it has to be mathematically tested and driven home. But here forces alone can not do the job - because each pair of 4 sides will exert a force in such a manner that the system starts rotating in clockwise or anticlockwise - just extend arrows of 4 sides with anti-clock. This type of system can be brought to equilibrium by 2 cross quadrupoles alone.

*** It is easier said than done - we cannot place 4 equal charges on 4 corner of a sqaure. If we start by placing on two adjacent corners then we have to keep (-6 C) inbetween before we carry in 3rd charge at L-symetry. Even then, 3rd charge cannot be kept bcause (-6C) will create perturbations.

So placing 2 charges at ends of 1 diagonal with (-6C) at centre is plausible option. Then carefully bring in two remaining charges at 2nd diagonal from opposite directions.

Answer 3

Consider any charge in the corner. The other 3 charges would create a force of repulsion. Let the corner charge be the one in the upper right corner q.

q2---------q1
|
|
q3---------q4

q1=q2=q3=q4
F=k(Q1 Q2)/R^2
k=8.99E+9 N m^2 / C^2

Total F= Fh +Fv + F(45) (vector equation!)

Ftotal vertical = F(1,4) + F(1,3)sin(45)
Ftv = kq^2[1/(1E-1)^2 + sin(45)1/(.01 +.01)]
Note 0.1^2= 0.01 :-)
Ftv = 8.99E+9 (3)^2[1/(1E-1)^2 + sin(45)1/(.01 +.01)]
F tv= 10,951,600,483, 290 = 10.95 E+12 N

F total horizontal = F(1,2) + F(1,3)cos(45)

Magnitude wise our horizontal and vertical components must be the same

So F th=10.95 E+12 N

Ftotal=sqrt( (F tv)^2 + (F th)^2)
Ftotal=F tv(sqrt(2))
Ftotal= 15.5 E+12N

Then from Coulomb’s Law Q= R^2F/(k q)
Where F=-Ftotal
R= sqrt(2) .1 m /2
R=.0707 m
Q=(.0707 )^2 x 15.5 E+12/(8.99E+9 x 3)
Q=2.87 C

I hope it helps.
Any questions?