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CAF123

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## Homework Statement

1) A small disc of radius R with surface normal ##\hat{k}## is placed a distance z from a point charge −q. Assuming that R ≪ z, derive an expression for the electric flux ##\Phi_E## passing through the disc.

2) Twenty-seven drops of salt water with the same radius are each charged to a potential of V volts. They are made to coalesce into a single drop. How does the potential of the new single drop compare with that of the original drops?

## The Attempt at a Solution

1)If R << z, then the disc will look like a point charge. So the E field a distance z away will be ##\vec{E} = -\frac{-q}{4\pi \epsilon_o z^2}\hat{k} = \frac{q}{4 \pi \epsilon_o z^2}\hat{k}##. (This result does not make sense - the E-field should be pointing in negative z, but the negative cancels?) EDIT: I see that the E-field points in -ve z

**because**the charge is negative, so I have essentially counted the same thing twice.

If I say that the disc is like a point charge, then how would I find the flux? For a disc, A = πR

^{2}, but I am unsure about how to apply this to the point charge.

2) I think this is simply 27V,since the potentials can be added like scalars (scalar summation). I don't know what else to say here. If the potential across the drop is V then one side of the drop is higher in potential than the opposite side by V volts? Then since potentials are scalar additive, I just add all the potential differences together and get 27V? Is that right? Maybe I should also say that the work required to put these drops together is W = 27(qV), q the charge on a single drop.

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