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Chapter 19: Problem 31

Two equipotential surfaces surround a \(+1.50 \times 10^{-8} \mathrm{C}\) point charge. How far is the \(190-\mathrm{V}\) surface from the \(75.0-\mathrm{V}\) surface?

Short Answer

Expert verified
The distance between the surfaces is approximately 1.09 meters.

Step by step solution

01

Understand the Concept

Equipotential surfaces are surfaces where the electric potential is constant. For a point charge, these surfaces are spheres centered around the charge. The potential difference between two such surfaces is given by the electric potential formula, which relates potential to charge and distance (radius from the charge).
02

Use the Potential Formula

The electric potential (V) from a point charge is given by the formula: \[ V = \frac{kQ}{r} \]where \( V \) is the electric potential, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) \), \( Q \) is the charge, and \( r \) is the distance from the charge.
03

Set Up Equations for Each Surface

For the first surface,\[ V_1 = \frac{kQ}{r_1} = 190 \, \text{V}\]For the second surface,\[ V_2 = \frac{kQ}{r_2} = 75 \, \text{V}\]
04

Express the Radii in Terms of V and Q

Re-arrange the potential formula to solve for \( r \):\[ r = \frac{kQ}{V} \]Substituting the known values into the expression for the two radii, we find:\[ r_1 = \frac{8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \times 1.50 \times 10^{-8} \, \text{C}}{190 \, \text{V}} \]and\[ r_2 = \frac{8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \times 1.50 \times 10^{-8} \, \text{C}}{75 \, \text{V}} \]
05

Calculate the Distances

Evaluate the expressions for \( r_1 \) and \( r_2 \):\[ r_1 \approx 0.7105 \, \text{m} \]\[ r_2 \approx 1.8018 \, \text{m} \]
06

Find the Difference

The distance between the two equipotential surfaces is the difference between \( r_2 \) and \( r_1 \):\[ \Delta r = r_2 - r_1 \approx 1.8018 \, \text{m} - 0.7105 \, \text{m} \approx 1.0913 \, \text{m} \]

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Key Concepts

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

Equipotential Surfaces
Equipotential surfaces are fascinating concepts in the field of electromagnetism. Imagine them as invisible shells or layers surrounding a point charge, where every point on a particular surface has the same electric potential.

What's special about them?
  • These surfaces are always perpendicular to electric field lines.
  • No work is done when moving a charge along an equipotential surface since the potential is constant.
  • In the case of a point charge, these surfaces are spherical.
Understanding these surfaces helps visualize how electric potential changes in space. For a point charge, moving from one equipotential surface to another involves a change in potential that's related to the distance from the charge.
Point Charge
A point charge is a simplified model of a charged particle, like an electron or proton, concentrating all its charge in a single point. This model helps us investigate and understand electric fields and potentials in a more straightforward way.

For a point charge:
  • The electric field radiates outward (if positive) or inward (if negative) from the charge.
  • The strength of the field diminishes with the square of the distance from the charge.
  • It serves as the foundation for calculating electric potential at a point in space.
In the exercise, the point charge is central to determining the equipotential surfaces, as its location dictates the geometry of the potential field around it.
Coulomb's Constant
Coulomb's constant, symbolized as \(k\), plays a critical role in electrostatics. It's used in the formula for calculating electric force and potential between charges. This constant has a value of \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\).

Here's why it's important:
  • It quantifies the relationship between electric force and distance.
  • It appears in the potential formula \(V = \frac{kQ}{r}\), linking potential with charge and distance.
  • It serves as a fundamental constant in many electrical calculations.
In our problem, Coulomb's constant helps compute the distances of the equipotential surfaces from the point charge, showing how potential dwindles as distance increases.

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Most popular questions from this chapter

Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is \(0 \mathrm{V} ?\)

Identical point charges of \(+1.7 \mu \mathrm{C}\) are fixed to diagonally opposite corners of a square. A third charge is then fixed at the center of the square, such that it causes the potentials at the empty corners to change signs without changing magnitudes. Find the sign and magnitude of the third charge.

What is the capacitance of a capacitor that stores \(4.3 \mu \mathrm{C}\) of charge on its plates when a voltage of \(1.5 \mathrm{V}\) is applied between them?

Two charges \(A\) and \(B\) are fixed in place, at different distances from a certain spot. At this spot the potentials due to the two charges are equal. Charge A is 0.18 \(\mathrm{m}\) from the spot, while charge \(\mathrm{B}\) is \(0.43 \mathrm{m}\) from it. Find the ratio \(q_{\mathrm{B}} / q_{\mathrm{A}}\) of the charges.

Two capacitors have the same plate separation, but one has square plates and the other has circular plates. The square plates are a length \(L\) on each side, and the diameter of the circular plate is \(L\). The capacitors have the same capacitance because they contain different dielectric materials. The dielectric constant of the material between the square plates has a value of \(\kappa_{\text {square }}=3.00 .\) What is the dielectric constant \(\kappa_{\text {circle }}\) of the material between the circular plates?
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