Question

Part A:

If a spherical raindrop of radius 0.700 mm carries a charge of -2.40 pC uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop.)

Part B:

Two identical raindrops, each with radius and charge specified in part (A), collide and merge into one larger raindrop. What is the radius of this larger drop, if its charge is uniformly distributed over its volume?

Part C:

What is the potential at its surface, if its charge is uniformly distributed over its volume?

Answer 1

Answer 2

Part A: To find the potential at the surface of the raindrop, we can use the formula for the electric potential due to a uniformly charged sphere at a point outside the sphere. The electric potential (V) at a distance r from the center of the sphere is given by:

V = k * (Q / R)

Where: k = Coulomb's constant ≈ 8.99 × 10^9 N m^2/C^2 Q = Total charge of the sphere R = Radius of the sphere

Given: Radius of the raindrop (r) = 0.700 mm = 0.700 × 10^-3 m Charge of the raindrop (Q) = -2.40 pC = -2.40 × 10^-12 C

First, we need to find the total charge (Q) of the raindrop. Since the charge is uniformly distributed over its volume, we can use the formula for the charge of a sphere:

Q = (4/3) * π * ε * R^3

Where: ε = Permittivity of free space ≈ 8.85 × 10^-12 C^2/(N m^2) R = Radius of the sphere

Substitute the given values to find Q:

Q = (4/3) * π * (8.85 × 10^-12) * (0.700 × 10^-3)^3 ≈ -1.176 × 10^-14 C

Now, calculate the potential at the surface of the raindrop (V) using the formula:

V = k * (Q / R)

V = (8.99 × 10^9) * (-1.176 × 10^-14) / (0.700 × 10^-3) ≈ -1.512 V

Therefore, the potential at the surface of the raindrop is approximately -1.512 volts.

Part B: When two identical raindrops merge into one larger raindrop, their charges combine. The total charge of the larger raindrop (Q_total) is the sum of the charges of the individual raindrops:

Q_total = 2 * (-1.176 × 10^-14) ≈ -2.352 × 10^-14 C

Now, we need to find the radius (R_total) of the larger raindrop. Since the charge is uniformly distributed over its volume, we can use the formula for the charge of a sphere mentioned above and solve for R:

Q_total = (4/3) * π * ε * R_total^3

R_total^3 = Q_total / ((4/3) * π * ε)

R_total^3 = (-2.352 × 10^-14) / ((4/3) * π * 8.85 × 10^-12)

R_total^3 ≈ -1.991 × 10^-6

R_total ≈ ∛(-1.991 × 10^-6) ≈ -0.0126 m

Since the radius cannot be negative, the larger raindrop's radius is approximately 0.0126 meters.

Part C: To find the potential at the surface of the larger raindrop, we can use the formula for electric potential as we did in Part A:

V_total = k * (Q_total / R_total)

V_total = (8.99 × 10^9) * (-2.352 × 10^-14) / 0.0126 ≈ -16.72 V

Therefore, the potential at the surface of the larger raindrop is approximately -16.72 volts.