Answer:
156250 N/C
Step-by-step explanation:
weight of drop, W = 2.5 x 10^-14 N
charge of drop, q = 1.6 x 10^-19 C
Let the electric field is E.
Weight is balanced by the electrostatic force
W = qE
2.5 x 10^-14 = 1.6 x 10^-19 x E
E = 156250 N/C
The electric field that will balance the weight of the oil drop can be calculated using the following:
electric force, F = e E ( where, e is the charge of an electron and E is the electric field)
weight, W = 2.5 ×10⁻¹⁴ N
e E = W
Substitute the values:
7.2m questions
9.5m answers