An oil drop of mass 3.25X10^-15 kg falls vertically with uniform velocity, through the air between vertical parallel plates which are 2 cm apart. When a p.d. of 1000V is applied to the plates ,the drop moves to the positively charged plate, being inclined at 45 degree to the vertical . Calculate the charge on the drop.

Let F be the resultant force on the charge when it is at 45° to the vertical. The horizontal component of this force, f = Fsin45° and the vertical component of this force f' = Fcos45°.

The vertical component of force equals the weight of the charge, mg.

So, f = mg

Fcos45° = mg (1) where m = mass of oil drop = 3.25 × 10⁻¹⁵ kg and g = acceleration due to gravity = 9.8 m/s²

Also, the horizontal component of the force equals the electric force F' = qE on the charge.

So, f' = qE

Fsin45° = qE (2) where q = charge on oil drop and E = electric field between the plates = V/d where V = potential difference between plates = 1000 V and d = distance between plates = 2 cm = 0.02 m

Dividing equation (2) by (1), we have

Fsin45°/Fcos45° = qE/mg

tan45° = qE/mg

tan45° = qV/dmg

Making q subject of the formula, we have

q = dmgtan45°/V

Substituting the values of the variables into the equation, we have

q = dmgtan45°/V

q = 0.02 m × 3.25 × 10⁻¹⁵ kg × 9.8 m/s² × tan45°/1000 V

q = 0.02 m × 3.25 × 10⁻¹⁵ kg × 9.8 m/s² × 1/1000 V

q = 0.637 × 10⁻¹⁸ C

q = 6.37 × 10⁻¹⁹ C

So, the charge on the oil drop is 6.37 × 10⁻¹⁹ C.

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