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Location \(A\) is \(3.00 \mathrm{m}\) to the right of a point charge \(q .\) Location \(B\) lies on the same line and is \(4.00 \mathrm{m}\) to the right of the charge. The potential difference between the two locations is \(V_{B}-V_{A}=45.0 \mathrm{V} .\) What are the magnitude and sign of the charge?

Short Answer

Expert verified
The charge is approximately \(-6.01 \times 10^{-8} \; \mathrm{C}\), negative.

Step by step solution

01

Understand the Potential Difference Formula

The electric potential difference between two points in the field of a point charge can be measured using the formula: \[ V_B - V_A = k_e \cdot \frac{q}{r_B} - k_e \cdot \frac{q}{r_A} \]where- \(q\) is the charge,- \(k_e = 8.99 \times 10^9 \; \mathrm{N \cdot m^2/C^2}\) is the Coulomb constant,- \(r_A = 3.00 \; \mathrm{m}\) and \(r_B = 4.00 \; \mathrm{m}\) are the distances from the charge to points A and B, respectively.
02

Substitute Known Values Into Equation

Substitute the given values and the distances into the potential difference formula:\[ 45.0 = 8.99 \times 10^9 \cdot \left( \frac{q}{4.00} - \frac{q}{3.00} \right) \]Simplify this into:\[ 45.0 = 8.99 \times 10^9 \cdot q \cdot \left(\frac{1}{4.00} - \frac{1}{3.00}\right) \]
03

Calculate the Difference in Fractions

Calculate the difference between the fractions:\[ \frac{1}{4.00} - \frac{1}{3.00} = \frac{3.00 - 4.00}{12.00} = -\frac{1}{12.00} \]
04

Solve for the Charge q

Now, solve for \(q\):\[ 45.0 = 8.99 \times 10^9 \cdot q \cdot \left(-\frac{1}{12.00}\right) \]Simplify and solve for \(q\):\[ q = \frac{45.0 \times (-12.00)}{8.99 \times 10^9} \]\[ q = \frac{-540.0}{8.99 \times 10^9} \]\[ q \approx -6.01 \times 10^{-8} \; \mathrm{C} \]
05

Determine the Sign of the Charge

Since the potential difference \(V_B - V_A\) is positive (45 V), and we found a negative result for \(q\), this means the charge is actually negative. A positive potential difference means the electric field is pointing towards lower potential, implying the source charge must be negative.

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Key Concepts

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

Point Charge
A point charge is a charge that is concentrated at a single point in space. It is an idealized concept, meaning in practical scenarios, actual charges are spread over a finite space, but for simplicity in calculations and understanding electric fields, we often treat them as point charges.

Understanding a point charge helps in calculating the electric field and potential at different locations. The electric field around a point charge decreases with the square of the distance from the charge, while the potential decreases with the distance. Imagine a point charge as a source of influence on the surrounding space; this influence diminishes as one moves further away from the charge.

In our given exercise, the charge influences two specific points (A and B) and creates a potential difference between them due to its existence alone.
Coulomb's Law
Coulomb's Law is fundamental to understanding electrostatic interactions. It describes the force between two point charges. The formula for Coulomb's Law is: \[ F = k_e \frac{|q_1 \, q_2|}{r^2} \]Here:
  • \(F\) is the force between the charges.
  • \(k_e\) is the Coulomb constant \(8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2}\).
  • \(q_1\) and \(q_2\) are the magnitudes of the charges.
  • \(r\) is the distance between the centers of the two charges.
This force can be attractive if the charges are of opposite signs or repulsive if they have the same sign.

In the context of electric potential, Coulomb's Law also assists in determining how point charges interact with each other and the surrounding electric field. In our problem, although it primarily focuses on electric potential, Coulomb’s Law principles ensure comprehending how the field interactions contribute to potential differences.
Electric Field
An electric field represents the influence a charge exerts on the surrounding space, affecting other charges that enter this space. It is a vector field, meaning that it has a direction and magnitude. The strength of the electric field \(E\) created by a point charge \(q\) is given by the formula:

\[ E = \frac{k_e \cdot q}{r^2} \]

where \(r\) is the distance from the point charge.

  • Direction: Outward from the charge if positive, inward if negative.
  • Units: Newtons per Coulomb (N/C).
In our exercise context, the negative sign of the charge means the electric field direction from this charge is inward, extending towards the charge itself. This behavior inversely influences the potential difference between points A and B, as the field dictates the rate at which the electric potential changes as you move along its direction.
Electric Potential Difference
Electric potential difference, also known simply as voltage, is a measure of the difference in electric potential energy per unit charge between two points. It shows how much energy work is done by moving a charge between two points in an electric field.

The relationship between the electric potential difference and a point charge’s potential at a distance \(r\) is described by the formula:\[ V = k_e \cdot \frac{q}{r} \]
  • \(V\): Electric potential (volts).
  • \(q\): The point charge creating the potential.
  • \(r\): The distance from the charge.
In the original exercise, the potential difference \(V_B - V_A = 45 \mathrm{V}\) tells us how energy per unit of charge changes as we move from point A to point B. The negative value of the charge along with this positive difference indicates that point A is at higher potential energy compared to point B, hence showing the direction of electric field and potential energy shift.

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Most popular questions from this chapter

Two identical point charges \(\left(q=+7.20 \times 10^{-6} \mathrm{C}\right)\) are fixed at diagonally opposite corners of a square with sides of length \(0.480 \mathrm{m}\). A test charge \(\left(q_{0}=-2.40 \times 10^{-8} \mathrm{C}\right),\) with a mass of \(6.60 \times 10^{-8} \mathrm{kg},\) is released from rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the square.

Refer to Multiple-Concept Example 3 to review the concepts that are needed here. A cordless electric shaver uses energy at a rate of 4.0 W from a rechargeable 1.5-V battery. Each of the charged particles that the battery delivers to the shaver carries a charge that has a magnitude of \(1.6 \times\) \(10^{-19} \mathrm{C} .\) A fully charged battery allows the shaver to be used for its maximum operation time, during which \(3.0 \times 10^{22}\) of the charged particles pass between the terminals of the battery as the shaver operates. What is the shaver's maximum operation time?

The same voltage is applied between the plates of two different capacitors. When used with capacitor A, this voltage causes the capacitor to store \(11 \mu \mathrm{C}\) of charge and \(5.0 \times 10^{-5} \mathrm{J}\) of energy. When used with capacitor \(\mathrm{B}\) which has a capacitance of \(6.7 \mu \mathrm{F}\), this voltage causes the capacitor to store a charge that has a magnitude of \(q_{\mathrm{B}} .\) Determine \(q_{\mathrm{B}}\).

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of \(0.75 \mathrm{mm} .\) When an electric spark jumps between them, the magnitude of the electric field is \(4.7 \times\) \(10^{7} \mathrm{V} / \mathrm{m} .\) What is the magnitude of the potential difference \(\Delta V\) between the conductors?

During a particular thunderstorm, the electric potential difference between a cloud and the ground is \(V_{\text {cloud }}-V_{\text {ground }}=1.3 \times 10^{8} \mathrm{V},\) with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?

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