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An oil drop of mass 3.25*10^(-15) kg falls vertically with uniform velocity , through the air between vertical parallel plates which are 2 m apart. When a p.d. of 1000 V is applied to the plates the drop moves to the positively charged plate , being inclined at 45 degrees to the vertical. Calculate the charge on the drop.
Asked by arjunsah797 | 15 Mar, 2022, 09:58: AM
Charged oil drop moves towards plate that is at +ve potential 1000 V , when it is subjected to force mg due to gravity and electrical force qE .

where m is mass of oil drop, g is acceleration due to gravity , q is magnitude of charge on oil drop and E is electric field intensity

If the direction of movement of oil drop makes 45o with vertical , then magnitudes of these two forces are equal .

qE = m g

magnitude of charge q = ( m g ) / E

Electric field intensity E = 1000 V / 2 m = 500 V/m

magnitude of charge q = ( 3.25 × 10-15 × 9.8 ) / 500 = 6.37 × 10-17 C

Since the charge is attracted towards +ve potential plate , sign of charge is negative

Hence , charge on oil drop = - 6.37 × 10-17 C
Answered by Thiyagarajan K | 15 Mar, 2022, 02:35: PM

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