Three point charges are fixed to the corners of a square, one to a corner, in such a way that the net electric field at the empty corner is zero. Do these charges all have (a) the same sign and (b) the same magnitude (but, possibly, different signs)? Which of the following are true?
1. Since the field at the empty corner due to the charge placed at the diagonally opposite corner will have equal x and y components, the charges placed at the other two corners must have the same magnitude.
2. It is certainly possible for all three occupied corners of the square to have charges with the same magnitude and still produce a zero electric field at the un-occupied corner of the square.
3. It is impossible to have any arrangement of charges, with any combination of sign and magnitude, at 3 of the 4 corners of a square such that the electric field at the un-occupied corner iszero.
4. The charge placed at the diagonally opposite corner to the empty corner must have a charge opposite in sign to the identical charges placed at the other occupied corners; and have a charge whose magnitude is larger than that of the other two charges.
5. The charges placed at the two corners along the edge of the square away from the empty must have the same sign. In that way the net field at the empty cornerdue to the two charges alone will lie along the diagonal of the square.
6. It is certainly possible for all three charges at the occupied corners of the square to have the same sign and still produce zero field at the un-occupied corner of the square.

1.- No

2.-No

3.-No

The arrangement below make it possible

4.-Yes

As describe in step by step explanation

5.-They need to have the same sign, because this is the only way you can compensate with other opposite charge

6.-No

Explanation:

We have to reasoning this way. Let say the empty corner is at the upper right of the square. if we call F1 and F2 forces acting in this point, due to charges of the left and down corners of the square respectivly. F1 will have direction of x axis and F2 will have direction of y axis. Now in order to cancel these two forces, we need:

1.-Charges in left  and down coners, must be of the same sign and magnitude

2.-the charge in diagonal opposite corner have to be of opposite sign and bigger charge than the two previous charges.

How big it need to be ??

E₁  = Ex  =  K*q₁/d²      E₂  =  Ey  = K*q₂/d²   (1)       q₁  =  q₂  =  q

distance between empty corner and its opposite corner is:

L = √ d² + d²   ⇒ L  = √2 * d

cos 45⁸ = sin 45⁸ = √2/2

E₃  ( field due to charge in opposite corner) is:

E₃ =K* q/ 2 * d²  

And  E₃ₓ  =  E₃y  =  cos 45⁸* E₃

E₃ₓ  = [ K*q/2 *d² ]* √2/2

E₃ₓ  =  K*q *√2/4*d²

Now if we compare   E₃ₓ   with   Ex  =  K*q₁/d²    equation (1)

q*√2/4  = q₁

then

q = 4*q₁/√2

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.