An oil drop whose mass is 1.2×10 15 kg is held at rest between two large plates
ID: 1611932 • Letter: A
Question
An oil drop whose mass is 1.2×1015 kg is held at rest between two large plates separated by 1.3 cm(Figure 1) when the potential difference between the plates is 330 V .
Part A
How many excess electrons does this drop have?
Express your answer as an integer.
An oil drop whose mass is 1.2×1015 kg is held at rest between two large plates separated by 1.3 cm(Figure 1) when the potential difference between the plates is 330 V .
Part A
How many excess electrons does this drop have?
Express your answer as an integer.
Explanation / Answer
As Oil drop is at rest between plates, Gravitational force (mg) acting in downward direction is balanced by electrical force ( Eq ) acting in upward direction.
E is electrcal field between plates and equal to V/d
where V is potential difference between plates and d is distance between plates.
q = ne ( where n is excess electrons carried by drop and e is magnitude of charge of electron = 1.6x10-19
Hence mg = V n e /d
n = mgd/ Ve
= 1.2x10-15 * 9.8* 0.013 * /330*1.6x10-19
= 3