Part THREE: Two small spheres each have a mass m of 0.30 g and are suspended as

Question

Part THREE:

Two small spheres each have a mass m of 0.30 g and are suspended as pendulums by light insulating strings from a common point, as shown below. The spheres are given the same electric charge, and the two come to equilibrium when each string is at an angle of =  7.0° with vertical. If each string is 2.0 m long, what is the magnitude of the charge on each sphere?

Part Four:

The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail. When small prey (or even potential mates) swim within a few feet of the fish, they perturb the electric field. The change in the field is picked up by electric sensor cells in the skin of the elephant nose. These remarkable fish can detect changes in the electric field as small as 3.0 N/C.

Explanation / Answer

Part Three:

To solve this question let us examine just one of the spheres(say left).In the free body diagram, there are 3 forces, mg down, electric repulsion left, and Tension at an angle.

Since the system is in equilllibrium, so in vertical direction,

Tcos = mg

Tcos(7 deg) = ( .0003 )( 9.8 )

T = 0.00296N

and in horizontal direction,

Fe = Tsin()

Fe = 0.00296 sin(7 deg)

Fe = 0.00036 N

Now, we use coloumbs law to find the values that provide this electric force

Fe = kq1q2/r2

Here r = 2* 2*sin(7 deg) = 0.487 m.

So,Fe = kq1q2/r2

0.00036 = (9*10^9)(q1)(q2)/(.487)2

However, q1 and q2 are identical as stated in the problem, so q X q = q2

0.00036 = (9*10^9)(q^2)/(.487)2

So e get q = 97.4 nC

Part Four:

As we know, E = kq/r2

3 X 10-6 = (9 X 109)(q)/(9*0.0254)2

q = 1.74*10^-17 C

Now the number of electrons,...

1.74*10^-17 / 1.6 X 10-19 = 4.79 electrons or 8 electrons.

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