If Capacitance does not depend on q then why do I use it to calculate the first part?

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anonymous · Answer

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anonymous · Answer

I find it kind of ironic that the book says that for (b) when the charge is increased the Capacitance remains the same because Capacitance is independent of q... yet we use q to calculate the first part... that's confusing.

anonymous · Answer

Is it because at first the objects are at rest and they cause an electric field between each other? Induced net charge without any new charge being apply to them?

anonymous · Answer

q is proportional to V (pot diff)
q = C * V,  C is a constant of proportionality

anonymous · Answer

so in 1st part, u have been given q and V
find C