To tackle this problem, we need to understand the concepts of electric field intensity, capacitance, and how they relate to charged spherical conductors. Let's break it down step by step.
Electric Field Intensity and Capacitance
The electric field intensity (E) just outside a charged spherical conductor can be expressed using the formula:
E = k * Q / r²
Where:
- E is the electric field intensity.
- k is Coulomb's constant, approximately 8.99 x 10^9 N m²/C².
- Q is the charge on the sphere.
- r is the radius of the sphere.
From this equation, we can rearrange it to find the charge (Q):
Q = E * r² / k
Calculating Charge for Each Drop
Let's calculate the charge for both mercury drops using their respective radii and the given electric field intensity of 9 x 10^9 N/C.
For the smaller drop (radius = 1 mm = 0.001 m):
Using the formula:
Q₁ = E * r₁² / k
Q₁ = (9 x 10^9 N/C) * (0.001 m)² / (8.99 x 10^9 N m²/C²)
Q₁ = (9 x 10^9) * (1 x 10^-6) / (8.99 x 10^9)
Q₁ ≈ 1.00 x 10^-6 C
For the larger drop (radius = 2 mm = 0.002 m):
Similarly, we calculate:
Q₂ = E * r₂² / k
Q₂ = (9 x 10^9 N/C) * (0.002 m)² / (8.99 x 10^9 N m²/C²)
Q₂ = (9 x 10^9) * (4 x 10^-6) / (8.99 x 10^9)
Q₂ ≈ 4.00 x 10^-6 C
Capacitance of Each Drop
The capacitance (C) of a spherical conductor is given by:
C = 4 * π * ε₀ * r
Where ε₀ (the permittivity of free space) is approximately 8.85 x 10^-12 F/m.
Capacitance of the Smaller Drop:
C₁ = 4 * π * ε₀ * r₁
C₁ = 4 * π * (8.85 x 10^-12 F/m) * (0.001 m)
C₁ ≈ 1.11 x 10^-13 F
Capacitance of the Larger Drop:
C₂ = 4 * π * ε₀ * r₂
C₂ = 4 * π * (8.85 x 10^-12 F/m) * (0.002 m)
C₂ ≈ 2.22 x 10^-13 F
Merging the Drops
When the two drops merge, they form a single spherical drop with a new radius (R) and charge (Q_total). The total charge is:
Q_total = Q₁ + Q₂
Q_total = 1.00 x 10^-6 C + 4.00 x 10^-6 C = 5.00 x 10^-6 C
The new radius (R) of the merged drop can be calculated using the volume conservation principle:
V = (4/3) * π * r³
Setting the volumes equal:
(4/3) * π * (0.001 m)³ + (4/3) * π * (0.002 m)³ = (4/3) * π * R³
Solving for R gives:
R³ = (0.001³ + 0.002³) = 1 x 10^-9 + 8 x 10^-9 = 9 x 10^-9
R = (9 x 10^-9)^(1/3) ≈ 0.00208 m
Potential on the Surface of the Merged Drop
The potential (V) on the surface of a charged sphere is given by:
V = k * Q_total / R
V = (8.99 x 10^9 N m²/C²) * (5.00 x 10^-6 C) / (0.00208 m)
V ≈ 2.16 x 10^4 V
Capacitance of the Merged Drop
Finally, the capacitance of the merged drop can be calculated as:
C_total = 4 * π * ε₀ * R
C_total = 4 * π * (8.85 x 10^-12 F/m) * (0.00208 m)
C_total ≈ 2.33 x 10^-13 F
In summary, we found the capacitance of each drop, calculated the potential and capacitance of the merged drop, and used fundamental principles of electrostatics to derive these results. If you have any further questions or need clarification on any of these steps, feel free to ask!
