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A total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius of 24.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) 48.0 cm; (b) 24.0 cm; (c) 12.0 cm.

Short Answer

Expert verified
(a) 65.54 V; (b) 131.88 V; (c) 131.88 V.

Step by step solution

01

Formula for Electric Potential

To find the electric potential due to a charged sphere, we use the formula for electric potential at a distance \( r \) from the center of a charged sphere: \[ V = \frac{kQ}{r} \] where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is the Coulomb's constant, \( Q = 3.50 \times 10^{-9} \, \text{C} \) is the charge, and \( r \) is the distance from the center of the sphere.
02

Electric Potential at 48.0 cm

For point (a) at a distance 48.0 cm (or 0.48 m):\[ V = \frac{8.99 \times 10^9 \times 3.50 \times 10^{-9}}{0.48} \] Calculating this gives us \( V \approx 65.54 \, \text{V} \).
03

Electric Potential on the Surface (24.0 cm)

For point (b) which is on the surface at a distance 24.0 cm (or 0.24 m):\[ V = \frac{8.99 \times 10^9 \times 3.50 \times 10^{-9}}{0.24} \] Calculating this gives us \( V \approx 131.88 \, \text{V} \).
04

Electric Potential Inside the Sphere (12.0 cm)

For point (c) inside the sphere at 12.0 cm (0.12 m). Since we are inside a conducting sphere, the potential remains constant and equal to the surface potential. So, \( V = 131.88 \, \text{V} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's constant
Coulomb's constant is a pivotal part of calculating the electric potential around charged objects. This constant, often denoted by \( k \), has a fixed value \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \). It helps to determine the force or potential generated by charges in a medium like air or vacuum.
Using this constant, you can effectively compute the interaction between point charges. This is done by incorporating it into formulas that relate distances and charges, like those used to find electric potential:
  • This constant simplifies calculations between electrostatic forces of point charges.
  • Ensures accuracy in the SI system of units.
  • Allows for comparisons across different equations involving electrostatics.
When dealing with spherical charge distributions, Coulomb's constant comes into play as it ties the charge and distance into the electric potential formula. Understanding this constant is crucial for solving problems where electric charge effects need quantification.
Charged Sphere
A charged sphere, especially a conducting one, influences the electric potential around and within it. In the context of electrostatics, a uniformly charged sphere has all its charge distributed over the surface. This is due to the nature of conductive materials where excess charges migrate to the outer surface.
When calculating the electric potential at a distance from the center of a charged sphere, it treats the entire charge as if it were concentrated at the center, thanks to the spherical symmetry.
This is key for two main scenarios:
  • Outside the Sphere: Electric potential, \( V \), decreases with increasing distance \( r \) and follows \( V = \frac{kQ}{r} \) where the point of interest is outside or on the surface.
  • Inside the Sphere: The potential remains constant and equals the potential at the surface since all charges reside on the surface.
Understanding how a charged sphere behaves aids in predicting voltages or potentials at various points relative to the sphere.
Electric Charge Distribution
Electric charge distribution can significantly affect electric potential and field generated by a structure such as a sphere. We often deal with uniform electric charge distributions, particularly for metal or conducting spheres, which means the charge is spread evenly over the sphere's surface.
In practice, steps involve understanding how this distribution affects calculations:
  • A uniform charge distribution allows simplification in calculations and predictions of electric potential and field lines.
  • Inside a conductor, charges rearrange themselves until equilibrium is reached, causing no electric field in the conductor's interior.
  • This phenomenon means the potential inside a conductor matches that of the surface.
For a charged sphere, this understanding is vital, especially when solving problems with internal and external points. It ensures you accurately predict potential values at given distances, hence resolving various electrostatic scenarios efficiently.

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Most popular questions from this chapter

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by 45.0 mm, and the potential difference between them is 360 V. (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+\)2.40 nC? (c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower. (d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

A metal sphere with radius \(R_1\) has a charge \(Q_1\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_2\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\)-direction. The potential difference between point \(a\) (at \(x =\) 0.60 m) and point \(b\) (at \(x =\) 0.90 m) is 240 V. (a) Which point, \(a\) or \(b\), is at the higher potential? (b) Calculate the value of \(E\). (c) A negative point charge \(q = -\)0.200 \(\mu\)C is moved from \(b\) to \(a\). Calculate the work done on the point charge by the electric field.

An object with charge \(q = -6.00 \times 10^{-9}\) C is placed in a region of uniform electric field and is released from rest at point \(A\). After the charge has moved to point \(B\), 0.500 m to the right, it has kinetic energy \(3.00 \times 10^{-7}\) J. (a) If the electric potential at point \(A\) is \(+\)30.0 V, what is the electric potential at point \(B\)? (b) What are the magnitude and direction of the electric field?

Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 cm in diameter, has mass 50.0 g, and contains \(-\)10.0 \(\mu\)C of charge. The other sphere is 40.0 cm in diameter, has mass 150.0 g, and contains \(-\)30.0 \(\mu\)C of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on them. (\(Hint:\) The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.)

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