KecitatiU answer if no relevant work shown 1) Two around what is considered your

Question

KecitatiU answer if no relevant work shown 1) Two around what is considered your finaManswer tsgmen for and charged near neutral metal negatively on each rod is brought sphere and compare the right spheres are in contact with each if the shown. A net charge other as sphere, then the left sphere is moved away. Give the sign of the the magnitudes of their charges. Remember to justify your answer. 2) Three point charges form an equilateral triangle as shown. Give the direction of the net force on the rightmost charge and justify your answer. 3) Three evenly-spaced point charges lie along a line as shown. Draw and label all forces acting on each charge and indicate the direction of the net force on each char (or state that there is no net force 2Q +Q +Q an has charge Q. The middle one has unknown Three evenly spaced point charges lie along a line. Each one on end three charges is zero charge and is equidistant from the other two. The net force on each of the 4) Does the unknown charge have the same sign as a or opposite sign? Justify your answer 5) Is the unknown charge greater than, less than, or equal to Q? Justify your answer the magnitude of 6) Express the unknown charge as a multiple of a keeping in mind your previous answers 7) A point charge Qu is at the origin. A second point charge Q2 is onthex-axis at x2 L, and a third point charge Q3 is on the x-axis at x3 3L. The net force on each charge is zero. Find the ratio Q2/Q1 (magnitude only) t CO two protons in a helium nucleus are separated by roughly 1 15m. If the strong force holding them together were to suddenly vanish, they would electrically repel each other. The resulting acceleration of each proton would be about 1.5 x m/s2. Find the missing power of ten. e 1.6x 10 c,ke 1 and 10 10? Score 30

Explanation / Answer

5) This problems calls for the use of the superposition principle,

Here we have two equally positively charged particles at the ends of the line. So we know that there must be a repulsive force acting on each of them.

Let us say that Force acting on the charge 1 due to charge 2 at the other end by coulomb's law is

F12= q2 /( 4 x pie x r 2)

Now we have been given the information in the question that this force be zero.

Net force becomes 2 F12, due to the forces F12 and F21.

So to cancel out these forces there should be some force acting in the opposite direction to these two forces and should cancel these two forces.

We can do this by keeping a negative charge of q/4 magintude in between these two positive charges.

F13 = - q2/ [4x(4 x pie x (r/2)2) ]= - F12

This force due to q/4 charge is thus canceling out the forces produced by the two positive charges of q magnitude.

Therefore the magnitude of the force to be kept between the two positive charges is - q/4 which is lesser in magnitude.

6) In the multiples it can be expressed as 0.25 q

7) Let us suppose that negative charge is present on the charge at x= L, and positive charges are present on x= 0 and x= 3L.

Computing magnitudes we get,

F13 = q1 q3/ (4 x pie x (3L)2 ) in + x direction

F31 = q1 q3 / ( 4 x pie x (3L)2) in -ve x direction

F12 = q1 q2 / ( 4 x Pie x (L)2 ) in + x direction

F23 = q1 q2 / (4 x pie x (2L) 2 ) in -ve x direction.

We know that since net force on each of the charges should be zero.

Th resultant force must be zero.

F13 + F31 = F12 + F23

On Solving we get, q2 / q3 = 8/ 45

Also on applying the balance on the Force blance on particle q3 ,

we get

F13 = F23

On solving we get,

q2/ q1 = 4/9 ( Answer)

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