An oil drop of mass 2×10^14 kg carries a charge of 8×10^-18c. the drop is stationary between two parallel plates 20mm apart with a potential difference of v between them. calculate v

Given data:

mass of oil  = 2 × 10¹⁴ kg,

charge (q)   = 8 × 10⁻¹⁸ C

drop is stationery between two parallel plates

and plates are at a distance of (d) = 20 mm

                                                        = 0.020 m,

Calculate potential difference (V) = ?

We know that the potential difference

                                  V = E.d Volts

                        Where

                                   d = distance between plates in meters

                                    E = electric field in N/C

                                      = m.g/q

                                    q = charge in coulombs

                                    m = mass in kg

Lets find out  

                        E = m.g/q  N/C

                           = (2 × 10¹⁴ × 9.8)/ 8×10⁻¹⁸    N/C

                           =  2.45 ×10³² N/C

 Now calculating V = E.d

                                 = 2.45 ×10³² × 0.020

                                 = 4.9  ×10³⁰ volts

The potential difference is 4.9  ×10³⁰ volts

the answer is b ) square root of 5