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Calculate the voltage needed to balance an oil drop carrying 10 electrons between the plates of capacitor 5mm apart.The mass of the drop is 3x10-16kg and g=10ms-2.

Atal Tiwari , 10 Years ago
Grade 12th pass
anser 3 Answers
Mallikarjun Maram

Last Activity: 10 Years ago

The gravitational force on the oil drop = mg = 3 x 10-16 x 10 = 3 x 10-15 N.
Magnitude of the electrostatic force needed balance this force is given by
E = mg/10e = (3 x 10-15) / (10 x 1.6 x 10-19) = 0.1875 x 104 V/m.
Voltage across the places of the capacitor is then given by
V = Ed = 0.1875 x 104 x 5 x 10-3 = 9.375 Volts.

Krish Gupta

Last Activity: 4 Years ago

 
The gravitational force on the oil drop  = mg = 3 x 10-16 x 10 = 3 x 10-15 N.
Magnitude of the electrostatic force needed balance this force is given by
E = mg/10e = (3 x 10-15) / (10 x 1.6 x 10-19) = 0.1875 x 104 V/m.
Voltage across the places of the capacitor is then given by
V = Ed = 0.1875 x 104 x 5 x 10-3 = 9.375 Volts.
Thanks , hope it helps.

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the solution to your problem.
 
For the oil drop to be in balance, the electrostatic force and the gravitational force need to be equal.
Hence, qE = mg
or, q x V/d = mg
or, V = mgd/q = mgd/10e
or, V = (3 x 10-16 x 10 x 5 x 10-3 )/ (10 x 1.6 x 10-19)
        = 9.375 Volts
 
Hope it helps.
Thanks and regards,
Kushagra

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