Question

A spherical water drop \(50.0 \mu \mathrm{m}\) in diameter has a uniformly distributed charge of \(+20.0 \mathrm{pC}\). Find (a) the potential at its surface and (b) the potential at its center.

Short answer

Answer: (a) The potential at the surface of the spherical water drop is 7.19 × 10³ V. (b) The potential at the center of the spherical water drop is also 7.19 × 10³ V.

Step-by-step solution

Recall Formula for Potential Due to a Point Charge

The electric potential V at a distance r from a point charge Q is given by the formula: \[V = \frac{kQ}{r}\] where k is the electrostatic constant equal to \(8.99 \times 10^9 Nm^2C^{-2}\).

Find the Charge on the Spherical Water Drop

The total charge is given as +20.0 pC. Converting the picocoulombs (pC) to coulombs (C): \[Q = 20.0 \times 10^{-12} C\]

Determine the Radius of the Sphere

We are given the diameter of the sphere as 50.0 µm. To find the radius r, divide the diameter by 2 and convert the micro-meters (µm) to meters (m): \[r = \frac{50.0 \times 10^{-6} m}{2} = 25.0 \times 10^{-6} m\]

Find the Potential at the Surface of the Sphere

By using the potential formula, we find the potential at the surface by replacing the distance r with the radius of the sphere: \[V_{surface} = \frac{kQ}{r} = \frac{8.99 \times 10^9 \times 20.0 \times 10^{-12}}{25.0 \times 10^{-6}}\] Calculate the value of the potential: \[V_{surface} = 7.19 \times 10^3 V\]

Find the Potential at the Center of the Sphere

Since the charge is uniformly distributed, the potential at the center of the sphere is equal to the potential at the surface of the sphere. This is because, at every point inside the completely uniformly charged sphere, the net electric field is zero, and since potential is related to electric field, they have the same value everywhere inside the sphere. Therefore, \[V_{center} = V_{surface} = 7.19 \times 10^3 V\] So, (a) the potential at the surface of the spherical water drop is 7.19 × 10³ V, and (b) the potential at the center of the spherical water drop is also 7.19 × 10³ V.