a) Two plane paralell conducting plates 15mm apart are held horizontal one above

Question

a) Two plane paralell conducting plates 15mm apart are held horizontal one above the other in air. The upper plate is maintained at a positive potential of 1500 V while the lower plate is earthed. Calculate the numner of electrons which must be attached to a small oil drop of mass 4.9x10^-15 kg if it remains stationary in the air between the plate (assume the density of the air is negligible in comparison to that of oil.)

b) if the potential of the upper plate is suddenly changed to -3500 V what is the initial acceleration of the charge drop?

Explanation / Answer

For oil drop to be stationary,

mg = q*V

charge = q = mg/V = 3.2x10-17C

no, of electron = q/e = q/(1.6x10-19C) = 200 electrons

If the upper plate is changed to -3500V suddenly,

then, ma = mg+qV2 = 4.8x10-14 + 3.2x10-17 * 3500 = 1.6x10-13

acceleration = 1.6x10-13 / 4.9x10-15 = 32.66m/s2 downwards

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