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At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V\(/\)m, respectively. (Take \(V = 0\) at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?

Short Answer

Expert verified
(a) The distance is 0.307 m. (b) The magnitude of the charge is approximately \(1.7 \times 10^{-10}\) C. (c) The electric field is directed away from the charge.

Step by step solution

01

Understanding the Relationships

Understand that the electric potential \( V \) and electric field \( E \) for a point charge \( Q \) at a distance \( r \) are given by \( V = \frac{kQ}{r} \) and \( E = \frac{kQ}{r^2} \) respectively, where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)). The goal is to find \( r \) and \( Q \).
02

Finding the Distance to the Point Charge

By dividing the formula for \( V \) by the formula for \( E \), we get: \( \frac{V}{E} = r \). Plug the known values into this equation: \( \frac{4.98}{16.2} = r \), yielding \( r = 0.307 \) m.
03

Calculating the Magnitude of the Charge

Using the value of \( r = 0.307 \) m and the formula for \( V \) (\( V = \frac{kQ}{r} \)), we solve for \( Q \). Rearrange to find \( Q = \frac{Vr}{k} \). Substituting the known values, \( Q = \frac{4.98 \times 0.307}{8.99 \times 10^9} \), resulting in \( Q \approx 1.7 \times 10^{-10} \) C.
04

Determining the Direction of the Electric Field

Since the electric field magnitude \( E \) and potential \( V \) are both positive, the point charge \( Q \) is positive. The electric field is directed away from positive charges.

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Key Concepts

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

Electric Potential
Electric potential is a scalar quantity that is associated with the electric field a charge produces. It is the amount of work needed to move a unit positive charge from a reference point (typically at infinity) to a specific point in the space around a charge, without any acceleration. This is expressed as the potential energy per unit charge.
For a point charge, the electric potential at a distance \( r \) from the charge \( Q \) is given by the formula:
  • \( V = \frac{kQ}{r} \)
where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \).
In the original problem, the potential \( V \) given is 4.98 V, suggesting the work done per unit charge to move it from infinity to the current position is 4.98 joules.
This scalar nature means that unlike vector quantities, potentials can be added algebraically without considering direction, making calculations straightforward.
Understanding electric potential is key to solving many electrostatics problems, especially those involving energy considerations.
Electric Field
The electric field is a vector field that represents the force experienced by a test charge at any point in space due to other charge distributions. It tells us the magnitude and direction of the force that would be felt by a unit positive charge placed at any given point.
For a point charge, the electric field \( E \) can be calculated using the equation:
  • \( E = \frac{kQ}{r^2} \)
An electric field magnitude of 16.2 V/m indicates the force per unit charge at the given distance from the charge is 16.2 newtons per coulomb.
It's vectorial nature makes the electric field more complex to deal with compared to scalar quantities like electric potential. It involves directional considerations. For example, the electric field direction is determined by the nature of the source charge; it points away from positive charges and towards negative ones.
Grasping the concept of electric fields aids in visualizing how charges influence their surroundings and helps solve problems related to forces and motion under electric influences.
Coulomb's Law
Coulomb's law is the foundational principle that describes how charged particles interact with one another. It mathematically defines the electric force between two point charges.
The law is represented by the equation:
  • \( F = \frac{k|Q_1 Q_2|}{r^2} \)
Here, \( F \) is the force between charges \( Q_1 \) and \( Q_2 \), \( r \) is the distance between the charges, and \( k \) is Coulomb's constant.
This concept is crucial for understanding electric interactions. It tells us:
  • The force is inversely proportional to the square of the distance, meaning it decreases rapidly with distance.
  • The force's magnitude is proportional to the product of the magnitudes of the charges involved.
  • Opposite charges attract, so the force is negative, while like charges repel, resulting in a positive force.
In the given problem, Coulomb's Law helps determine the direction and magnitude of the forces and fields created by point charges, providing insights into how charges behave and influence one another in electrostatic situations.

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Most popular questions from this chapter

A ring of diameter 8.00 cm is fixed in place and carries a charge of \(+\)5.00 \(\mu\)C uniformly spread over its circumference. (a) How much work does it take to move a tiny \(+\)3.00-\(\mu\)C charged ball of mass 1.50 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 maximum speed it will reach?

Charge \(Q = +\)4.00 \(\mu\)C is distributed uniformly over the volume of an insulating sphere that has radius \(R =\) 5.00 cm. What is the potential difference between the center of the sphere and the surface of the sphere?

(a) How much work would it take to push two protons very slowly from a separation of \(2.00 \times 10^{-10}\) m (a typical atomic distance) to \(3.00 \times 10^{-15}\) m (a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

A very large plastic sheet carries a uniform charge density of \(-\)6.00 nC\(/\)m\(^2\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by 1.00 V. What type of surfaces are these?

Two point charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two point charges having opposite signs.

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