Question

Solve numerical example.

A potential difference of 5000 volts is applied between two parallel plates 5cm apart a small oil drop having a charge of 9.6 ×10^-19 C falls between the plates. Find

1. electric field intensity between the plates and
2. the force on the oil drop.

Answer 1

 Given: V = 5000 volts, d = 5 cm = 5 × 10^-2 m, q = 9.6 × 10^-19 C

To find: a. Electric field intensity (E)
b. Force (F)

Formulae: i. E = `"V"/"d"`

ii. E = `"F"/"q"`

Calculation: From formula (i),

E = `5000/(5xx10^-2)` = 10⁵ N/C

From formula (ii)
F = E × q
= 10⁵ × 9.6 × 10^-19
= 9.6 × 10^-14 N

1. Electric field intensity between the plates is 10⁵ N/C
2. Force on the oil drop is 9.6 × 10^-14 N