Three tiny charged metal balls are arranged on a straight line. The middle ball

Question

Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls are negatively charged. The two outside balls are separated by 20 cm and the middle ball is exactly halfway in between. (HINT: Draw a picture; in your picture, the distance between the two outermost balls should be 20 cm.) The absolute value of the charge on each ball is the same, 1.75 Coulombs (the meaning of , which is read as "micro", is 10-6). Give your answers in newtons.

(a) What is the magnitude of the attractive force on either outside ball due ONLY to the positively-charged middle ball?

(b) What is the magnitude of the repulsive force on either outside ball due ONLY to the other outside ball?

(c) What is the magnitude of the net force on either outside ball?

Explanation / Answer

Let the three charges be q1 , q2 , and q3

Let q2 be the middle and positive charge

Let q1 and q3 be outside negative charges

Then q1 = q2 = q3 = 1.75 µC = 1.75 * 10-6 C

Let the distance b/w the two outside balls be r1 = 20 cm = 0.2 m

Since q2 is in the middle , its distance from either outside charges be r2 = 10 cm = 0.1 m

From the principle of superposition , individual forces are unaffected by the presence of other forces

So the force b/w any two charge can be easily calculated by using coulomb’s law

a)The force b/w the charges q2 ( positive ) and q1 or q3 ( negative )

We can calculate the magnitude of the force F21 b/w q1 and q2 and this will be equal to the force b/w q2 and q3

F21 = 1/40 * q3 * q2/( r2 )2

        = 9 * 109 * ( 1.75 * 10-6 * 1.75 * 10-6 )/( 0.1 )2

         = 27.56 * 10-3/0.01

         = 2756.25 * 10-3 C = 2.756 C

So the magnitude of the force b/w the the positive middle charge and negative outside charge is F21 or F23 = 2.756 C

b)Force b/w the two outside charges q1 and q3

Let the force b/w q1 and q3 be F13

Then F13 = 1/40 * q1 * q3 / ( r1 )2

                = 9 * 109 * 1.75 * 1.75 * 10-6 * 10-6/ ( 0.2 )2

                = 27.56 * 10-3/( 0.2 * 0.2 )

                = 689.6 * 10-3 C

                 = 0.689 C

So the magnitude of the force b/w the outside negative charges is F13 = 0.689 C

c)The net force acting on either of the outside charges

The net forces on both outside charges is equal.So let us calculate the net force on q3

The net force acting on the outside charge is simply the vector sum of the force due to charge q1 and force due to q3

Fnet = F23 + F13 = F23 – F13   ( since force b/w outside charges is repulsive F13 is negative )

       = 2.756C – 0.689 C = 2.067C

So the total force on the outside charges is 2.067C and its direction is towards the centre

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