An oil drop of mass 3.2 ×10-15 kg falls vertically with uniform velocity through the air between vertical parallel plates which are 2cm apart. When a p.d. of 1000V is applied to the plates the drop moves to the negatively charged plate, its path being inclined at 45° to the vertical. Calculate the charge on the drop.
Bijay_Magar Nice question brother.
Fe = mg
QE = mg [Fe = QE]
Q = (mg)/E
Q = (mg)/(V/d) [E = V/d ]
Q = mgd/V
Q = ( 3.2×10-5 x 9.8 × 0.2 )/1000
Q = 6.272 × 10-8
Don’t hesitate to correct me 😀.