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Can anyone demonstrate how to get the answers to these questions? I'm just interested in the method I need to use in order to obtain the correct answer no matter what the values are.

Three small spheres are placed at fixed points along the x-axis, whose positive direction points towards the right.

Sphere A is at x = 36.0 cm, with a charge of –8.00 μC.

Sphere B is at x = 41.0 cm, with a charge of 9.00 μC.

Sphere C is at x = 46.0 cm, with a charge of –3.00 μC.

a) Calculate the magnitude of the electrostatic force on sphere B.

b) Sphere B is now removed: What would be the magnitude of the electric field at the point where sphere B was located?

c) Sphere B is still missing. Give the x-coordinate of the point on the x-axis where the field due to spheres A and C is zero

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closed as off-topic by Emilio Pisanty, akhmeteli, Qmechanic Oct 25 '13 at 10:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

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The spheres are effectively point charges, so the force between any pair of spheres is just given by Coulomb's law. The forces are additive so in (a) the net force on $B$ is just the vector sum of $F_{AB}$ and $F_{BC}$.

The field strength is the force (from Coulomb's law) on a point charge of 1 Coulomb, so in q (b) proceed as before replacing $B$ by a charge of 1 Coulomb.

(c) is just the point between $A$ and $C$ where the force on a test charge sums to zero.

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Since the charges are distributed on spheres they can be considered as point charges.

Both the charges A and C will attract charge B so the force is given by subtracting these two forces vectorially

The electrostatic field due to A and C at point of B comes out to be 18 × 10^6 towards A.

Correspondingly the force experienced by B would be 162N towards A.

The field due to both charges would be zero at 6.2 cm from A towards B which would be 42.2 cm.

Now the interesting part is that even if we suppose that the charges are held together by some unknown uninterferring forces we do not bother to think about the induction i myself was too much bothered by this till sometime back when i read an article that the induction that could actually change your answer and that too drastically comes at extremely close distances. Normally it is just negligible, i would encourage you to read this articld here as it would clarify even more that why I could solve your question so easily without the need of some supercomputer to account for the induction.

http://www.nature.com/news/like-attracts-like-1.10698

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