Question

A charged oil drop of mass m = 5x10⁻³ kg is stationary (at rest) above a uniformly charged infinite plane sheet that carries positive charge density σ = +8.85nC/m². The charge (in uc) of the oil drop is:

A) 0 μC
B) +50 μC
C) -50 μC
D) -100 μC

Answer 1

The charge of the oil drop can be calculated by setting the gravitational force equal to the electric force using the formula mdrop × g = q × E. By calculating E based on the given charge density and using the mass of the drop, the charge q in microcoulombs (uc) can be determined.

Step-by-step explanation:

The charge (in uc) of the oil drop can be determined by using the equilibrium condition where the gravitational force is balanced by the electric force. The formula that relates these forces is mdrop × g = q × E, where mdrop is the mass of the oil drop, g is the gravitational acceleration (9.8 m/s²), q is the net charge of the oil drop, and E is the electric field due to the charged plane. Millikan's experiment provides a way of measuring the charge of individual oil drops using this principle.

Since we need to find the charge q, we can rearrange the equation to q = (mdrop × g) / E. However, we must first calculate E, the electric field, using the charge density σ of the plane and the equation E = σ / ε₀, where ε₀ is the vacuum permittivity (8.85 × 10⁻¹² C²/(N·m²)).

To find q, we will insert the mass of the oil drop and the calculated electric field into the equation, and solve for q. Converting this charge into microcoulombs (uc) will give us the answer.