Question

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

Answer 1

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