Two identical conducting small spheres are placed with their...

Two identical conducting small spheres are placed with their centers 0.300 $\mathrm{m}$ apart. One is given a charge of 12.0 $\mathrm{nC}$ and the other a charge of $-18.0 \mathrm{nC}$ . (a) Find the electric force exerted by one sphere on the other. (b) What If? The spheres are connected by a conducting wire. Find the electric force between the two after they have come to equilibrium.

Key Concepts

Coulomb's Law

Coulomb's Law is used to calculate the electric force between two point charges. It states that the magnitude of the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This law is essential when analyzing forces between charged objects, as it provides a method to determine both the direction and the magnitude of the electric interaction based solely on the values of the charges and their separation.

Charge Redistribution in Conductors

When conductors are connected by a wire, they reach an equilibrium state where the electric potential is uniform across the entire conductor system. This process involves the redistribution of charge between the conductors until they have equal potential, often resulting in charges being shared equally if the conductors are identical. Understanding this phenomenon is crucial for analyzing changes in electric forces after connectors alter the charge distribution on initially isolated charged objects.

Transcript

00:02 So for the part of this problem, we need to find the pure attractive electrostatic force, we will see that it is attractive from the sign from the general formula k over times q1, q2 over the radial distance squared.

00:28 Substituting the node values you obtain that this is 9 .10 to the 9th power, then to the 9th power, then to times 10 the negative 9 power times minus 18 times 10 to the negative 19 power over 0 .3 squared and this is in newtons.

01:12 So when we put this in the calculator we obtain that f equals 2 .16 times 10 to the negative 15 power newton's and the sign is plus plus so the thought this this force is going to be attractive force for this problem this is also due to fact that the the sign of one sphere is negative and the sign of another sphere is positive so the total force is attractive and now for the part b for the part b we need to see when the spheres are connected so when we connect charged bodies generally charges will flow so your charges i connect charges i connect them this connection allows the flow of charges so if one is negative 18 times 10 to the minus 19 columns charges will flow until the potential on both sides is equal so this is on 12 times 10 the minus 90 power and clearly we see that the positive potential is much greater than the negative.

03:20 The positive charges are in large greater number of negative charges.

03:24 So what will happen is that positive charges will flow, but this is only in quotation flow.

03:34 Positive charges are, they cannot move, and only negative charges, move, electrons move.

03:39 So electrons will move here until both sides are at the same potential.

03:46 So we should start from the net charge, which is q net n -n equals 12.

03:54 Nanoculans and also this is 19 this is to the 19 power to the 9th power not 19th power to the 9th power and i have a typo here so this is to the 9th power since nano -coolins are on minus 9 this was using the calculation so the net charge is equal to 12 nano columns plus negative 18 negative nanocolums and this equals to negative six nanocombs.

05:00 So the charge, since since potentials cannot be equalized then there will be the same charge on each sphere and that charge will be one half of the net charge which is negative three nanomono columns...

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