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Two positive point charges, each of magnitude \(q,\) are fixed on the \(y\) -axis at the points \(y=+a\) and \(y=-a\) . Take the potential to be zero at an infinite distance from the charges. (a) Show the positions of the charges in a diagram. (b) What is the potenntial \(V_{0}\) at the origin? (c) Show that the potential at any point on the \(x\) -axis is $$ V=\frac{1}{4 \pi \epsilon_{0}} \frac{2 q}{\sqrt{a^{2}+x^{2}}} $$ (d) Graph the potential on the \(x\) -axis as a function of \(x\) over the range from \(x=-4 a\) to \(x=+4 a\) . (e) What is the potential when \(x \gg a ?\) Explain why this result is obtained.

Short Answer

Expert verified
Potential at the origin is \( \frac{1}{2\pi\epsilon_0} \cdot \frac{q}{a} \). On the x-axis, potential is \( \frac{1}{4\pi\epsilon_0} \cdot \frac{2q}{\sqrt{a^2+x^2}} \). For \( x \gg a \), potential approximates \( \frac{1}{4\pi\epsilon_0} \frac{2q}{x} \).

Step by step solution

01

Diagram the Charge Configuration

The two positive point charges, each of magnitude \( q \), are located on the \( y \)-axis, one at \( y = +a \) and the other at \( y = -a \). Draw the \( y \)-axis with these two charges marked at their respective positions.
02

Calculate Potential at Origin

Since both charges are equidistant from the origin, the potential at the origin is the sum of the potentials due to each charge. The distance from the origin to each charge at \( y = a \) or \( y = -a \) is \( a \). Thus, the potential, \( V_0 \), is twice the potential from one charge, which is given by:\[ V_0 = 2 \times \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{a} = \frac{1}{2\pi\epsilon_0} \cdot \frac{q}{a} \].
03

Derive Potential on the x-axis

Consider a point on the \( x \)-axis at distance \( x \) from the origin. The distance from this point to each charge is \( \sqrt{a^2 + x^2} \). Potential from each charge at this point is \( \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{\sqrt{a^2 + x^2}} \). Therefore, the total potential is:\[ V = 2 \times \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{\sqrt{a^2 + x^2}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{2q}{\sqrt{a^2 + x^2}} \].
04

Graph the Potential on the x-axis

Graph the expression derived for \( V \) in Step 3 over the range \( x = -4a \) to \( x = 4a \). The plot should show that the potential decreases as \( |x| \) increases due to the increasing distance from the charges.
05

Calculate Potential for x ≫ a

When \( x \gg a \), the term \( \sqrt{a^2 + x^2} \approx x \) as \( a^2 \) becomes negligible. Hence, the potential simplifies to:\[ V \approx \frac{1}{4\pi\epsilon_0} \cdot \frac{2q}{x} \]. This simplification shows that at large distances, the system behaves like a single point charge located at the midpoint of \( y = 0 \).

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Key Concepts

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

Point Charges
In electrostatics, point charges are idealized charges with negligible size. They are often used to simplify complex charge distributions, allowing us to focus on the mathematical concepts without needing to worry about the physical size. In this exercise, we have two positive point charges, each with magnitude \( q \).

These charges are situated symmetrically along the \( y \)-axis, one at \( y = +a \) and the other at \( y = -a \). This configuration allows us to analyze potential as a function of different positions on the \( x \)-axis. Understanding how point charges behave is crucial, as it helps us predict electric forces and potentials in an electrical field.
  • Point charges are theoretical constructs with all charge concentrated at a single point.
  • They are used to simplify calculations in electric field and potential problems.
  • In practice, they approximate the behavior of small charged objects at distances large compared to their size.
The symmetrical placement on the \( y \)-axis provides a basis for calculating potential at various points, leveraging the principles of superposition in electrostatics.
Electric Potential Energy
Electric potential energy is a fundamental concept defining the work needed to move a charge within an electric field. For point charges, this energy depends on the positions of the charges and their magnitudes. The potential at a point is the electric potential energy per unit charge at that location.

In this problem, we explore the potential energy at the origin and along the \( x \)-axis. The potential at the origin is the sum of potentials due to each point charge, given their symmetric placement around the origin. This results in a simplified equation that considers the distances and magnitudes of these charges:
  • The total potential energy at any point is due to the cumulative effect of both point charges.
  • The principle of superposition simplifies calculations by allowing potentials to be added algebraically.
  • At the origin, potential energy sums due to equal distances from each charge, simplifying calculations.
Potential at any point on the \( x \)-axis further defines how distance affects potential energy, essential for understanding electric fields.
Distance on x-axis
The distance on the \( x \)-axis is a variable denoting positions of interest in relation to the point charges placed along the \( y \)-axis. This distance is critical in the formula for electric potential, particularly when examining how potential changes across different points.

For any position \( x \) on the \( x \)-axis, the distance to each charge is calculated using the Pythagorean theorem as \( \sqrt{a^2 + x^2} \). This diagonal distance considers both \( x \) and the fixed \( y \)-axis positions of the charges. It reveals how potential decreases with increased \( x \), particularly when \( x \) greatly exceeds \( a \), simplifying the formula to resemble that of a lone point charge potential.
  • Distance affects how the potential felt at a point on the \( x \)-axis is calculated.
  • At large distances (\( x \gg a \)), the system behaves like a single charge, indicating a monopole effect.
  • The algebraic expression for potential becomes approximate due to negligible \( a^2 \).
This concept underlines the importance of positioning and its relation to the calculated potential, crucial for modeling electric fields in various contexts.

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Most popular questions from this chapter

A total electric charge of 3.50 \(\mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of \(24.0 \mathrm{cm} .\) 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) \(48.0 \mathrm{cm} ;(\mathrm{b}) 24.0 \mathrm{cm}\) (c) \(12.0 \mathrm{cm} .\)

(a) If a spherical raindrop of radius 0.650 \(\mathrm{mm}\) carries a charge of \(-1.20 \mathrm{pC}\) uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop.) (b) Two identical raindrops, each with radius and charge specified in part (a), collide and merge into one larger raindrop. What is the radius of this larger drop, and what is the potential at its surface, if its charge is uniformly distributed over its volume?

A potential difference of 480 \(\mathrm{V}\) is established between large, parallel, metal plates. Let the potential of one plate be 480 \(\mathrm{V}\) and the other be 0 \(\mathrm{V}\) . The plates are separated by \(d=1.70 \mathrm{cm} .\) (a) Sketch the equipotential surfaces that correspond to \(0,120,\) \(240,360,\) and 480 \(\mathrm{V}\) . (b) In your sketch, show the electric field lines. Does your sketch confirm that the field lines and equipotential surfaces are mutually perpendicular?

A ring of diameter 8.00 \(\mathrm{cm}\) is fixed in place and carries a charge of \(+5.00 \mu \mathrm{C}\) unifurnly spread over its circumference. (a) How much work does it take to move a tiny \(+3.00-\mu \mathrm{C}\) charged ball of mass 1.50 \(\mathrm{g}\) from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maxi- mum speed it will reach?

A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a,\) and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{J}\) . When the second charge is moved to point \(b\) , the electric force on the charge does \(-1.9 \times 10^{-8} \mathrm{J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b ?\)

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