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

Expert verified
Answer: The potential at both the surface and the center of the water drop is 7192 volts.

Step by step solution

01

Identify given values

We are given the diameter (\(50.0 \mu m\)) of the water drop, and its charge (\(+20.0 pC\)). Let's convert these values to meters and Columbs respectively: Diameter = \(50.0 \mu m = 50.0 × 10^{-6} m\) Charge = \(+20.0 pC = 20.0 × 10^{-12} C\)
02

Find the radius of the water drop

To find the radius of the water drop, divide the diameter in meters by 2: Radius = \(\frac{50.0 × 10^{-6} m}{2} = 25.0 × 10^{-6} m\)
03

Calculate potential at the surface

Use the formula for potential to find the potential at the surface of the charged sphere: \(V_{surface} = \frac{kQ}{r}\) \(V_{surface} = \frac{8.99 × 10^9 Nm^2 C^{-2} × 20.0 × 10^{-12} C}{25.0 × 10^{-6} m}\) \(V_{surface} = 7192 V\) The potential at the surface of the water drop is 7192 volts.
04

Calculate potential at the center

For a uniformly charged sphere, the potential at the center is equal to the potential at the surface (since the electric field inside is zero). So the potential at the center is: \(V_{center} = V_{surface}\) \(V_{center} = 7192 V\) The potential at the center of the water drop is also 7192 volts.

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