Consider a horizontal parallel plate capactor where the size of the plates is mu

Question

Consider a horizontal parallel plate capactor where the size of the plates is much larger than the spacing between the plates. Between the plates is placed a small charged bead with a mass 2.5x10-12kg and a charge equal to that of 800 electrons. The separation between the plates is 4.1 cm and the magnitude of the charge on the two plates of the capacitor is equal, the top being positive and the bottom negative.

If the charge on the plates is adjusted until the bead is stationary (doesn't fall due to gravity), what is the resulting electrostatic potential difference between the plates?

Give you answer in kV (103V), so that if your answer is 450,000 V, you should enter 450.

Give your results to three digits and do not include the units in your answer.

Explanation / Answer

weight = m g

electric force = q E

E = V/d

weight = electric force

==> m g = q E = q V/d

==> V = d m g/q = 4.1e-2*2.5e-12*9.8/(800*1.6e-19) = 7.85 kV

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