Chapter 23: Problem 37
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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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Spherical Charge Distribution
In physics, understanding how charges are distributed in a sphere is crucial for solving many problems related to electrostatics. A spherical charge distribution refers to charges spread evenly across the surface or volume of a sphere.
This kind of distribution is particularly important when dealing with objects like droplets, planets, or even atomic nuclei where the shape is nearly spherical.
For a spherical charge, we often assume that the charge is distributed uniformly, meaning each part of the sphere holds the same amount of charge in relation to its size. This makes calculations more manageable. The symmetry of the shape allows us to use special formulas for calculating electric fields and potentials. Understanding this distribution helps in solving complex problems by simplifying them into more straightforward calculations.
This kind of distribution is particularly important when dealing with objects like droplets, planets, or even atomic nuclei where the shape is nearly spherical.
For a spherical charge, we often assume that the charge is distributed uniformly, meaning each part of the sphere holds the same amount of charge in relation to its size. This makes calculations more manageable. The symmetry of the shape allows us to use special formulas for calculating electric fields and potentials. Understanding this distribution helps in solving complex problems by simplifying them into more straightforward calculations.
Uniform Charge Distribution
A uniform charge distribution means that the charge is spread consistently throughout the object. In the case of a spherical water drop, this means that every part of the sphere's surface holds the same amount of charge per unit area. This is important because it allows us to easily calculate electric potentials and fields.
Such simplifications make it possible to apply formulas directly without worrying about complex variations in charge within the object.
- This kind of distribution assumes no variation in density, so the charge is evenly spread.
- Because of this uniform distribution, the sphere can be considered as if the entire charge is concentrated at its center for external points.
Such simplifications make it possible to apply formulas directly without worrying about complex variations in charge within the object.
Electric Potential Calculation
Calculating the electric potential is an essential step when dealing with charged objects. The electric potential indicates how much work is needed to move a charge within an electric field. For a spherical charge, the formula used is akin to that of a point charge.
The formula for the potential at the surface of a charge sphere is: \[ V = \frac{kQ}{r} \] where:
In this problem, since the electric field inside the sphere is zero, the potential at the center is the same as at the surface. This simplifies the calculations significantly, often leading to the same numerical result for both locations.
The formula for the potential at the surface of a charge sphere is: \[ V = \frac{kQ}{r} \] where:
- \( V \) is the potential,
- \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \),
- \( Q \) is the total charge,
- \( r \) is the radius.
In this problem, since the electric field inside the sphere is zero, the potential at the center is the same as at the surface. This simplifies the calculations significantly, often leading to the same numerical result for both locations.
Physics Problem Solving
Solving physics problems systematically can make even complex problems straightforward. To solve an electrostatic problem, it's essential to follow a structured approach. Start by identifying what is given and what needs to be found.
In our specific problem, we had to find the electric potential at the surface and the center of a charged sphere. Recognizing that the electric field inside a uniformly charged sphere is zero helps us understand why the potential at the center remains constant. These steps involve using physical principles to simplify what initially looks like a challenging problem.
- First, convert all the given quantities to standard units, such as meters and coulombs.
- Second, determine the shape and distribution of charges.
- Third, apply relevant formulas based on this shape and distribution.
In our specific problem, we had to find the electric potential at the surface and the center of a charged sphere. Recognizing that the electric field inside a uniformly charged sphere is zero helps us understand why the potential at the center remains constant. These steps involve using physical principles to simplify what initially looks like a challenging problem.