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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 \(12.0 \mathrm{V} / \mathrm{m},\) respectively. (Take the potential to be zero 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) 0.415 m, (b) \(2.30 \times 10^{-10} \mathrm{C}\), (c) Electric field is directed away from the charge.

Step by step solution

01

Relate Electric Field and Potential

For a point charge, the potential \( V \) at a distance \( r \) is given by \( V = \frac{kQ}{r} \) and the electric field \( E \) is given by \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \)). We can express \( Q \) in terms of the electric field: \( E = \frac{V}{r} \).
02

Calculate the Distance "r" from the Point Charge

Substitute the given values \( V = 4.98 \ \mathrm{V} \) and \( E = 12.0 \ \mathrm{V}/\mathrm{m} \) into the equation \( E = \frac{V}{r} \).\[ r = \frac{V}{E} = \frac{4.98}{12.0} = 0.415 \ \mathrm{m} \]
03

Determine the Magnitude of the Charge "Q"

Using the expression \( V = \frac{kQ}{r} \), solve for \( Q \):\[ Q = \frac{Vr}{k} = \frac{4.98 \times 0.415}{8.99 \times 10^9} \approx 2.30 \times 10^{-10} \ \mathrm{C} \]
04

Assess the Electric Field Direction

Since the potential is positive and assuming the charge is positive, the electric field would point away from the charge. The electric field direction is away from a positive charge and toward a negative charge.

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Key Concepts

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

Electric Potential
In the realm of electrostatics, electric potential is a fundamental concept. It represents the potential energy per unit charge at a specified point in an electric field. This means how much work is required to move a unit charge from infinity to that point, without accelerating the charge.
For a point charge, the electric potential, denoted as \( V \), can be determined using the formula:
  • \( V = \frac{kQ}{r} \)
where:
  • \( k \) is the Coulomb's constant \( (8.99 \times 10^9 \, \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2) \)
  • \( Q \) is the magnitude of the charge
  • \( r \) is the distance from the point charge
By setting the potential to zero at infinity, we define a reference point that simplifies calculations and comparisons. Here, the electric potential is directly proportional to the charge magnitude and inversely proportional to the distance from the charge. Thus, if you're closer to the charge, the potential increases.
Coulomb's Law
Coulomb's Law offers a detailed explanation of how forces between charged particles operate. It specifically shows how two point charges exert forces on each other, depending on their magnitudes and the distance separating them.
Coulomb's Law is mathematically expressed as:
  • \( F = \frac{k |Q_1 Q_2|}{r^2} \)
where:
  • \( F \) is the electrostatic force
  • \( Q_1 \) and \( Q_2 \) are the magnitudes of the two charges
  • \( r \) is the distance between the centers of the two charges
This law helps us understand that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Importantly, this tells us that as charges come closer, the force increases significantly due to the square of the distance term. In the context of our problem, Coulomb's Law underlies the calculations related to the electric field and potential.
Electric Field Direction
Analyzing the direction of an electric field provides insights into how charges will behave within that field. The electric field direction is defined as the direction a positive test charge would move if placed within the field.
In simple terms:
  • From a positive charge, the electric field lines radiate outward.
  • Toward a negative charge, the electric field lines converge.
This directional behavior helps us predict charge interactions. For instance, when we're told the potential is positive, and if we assume the charge is positive, the field must be moving away from the charge. This fundamental understanding of direction is key in solving various electrostatic problems, such as finding how an unknown charge relates to its surrounding space, as in the initial problem statement. Understanding this, we can infer the electric field direction based on the known charge polarity and the nature of electric field lines.

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Most popular questions from this chapter

A 5.00 pF parallel-plate air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to \(1.00 \times 10^{2} \mathrm{V}\) . The electric field between he plates is to be no greater than \(1.00 \times 10^{4} \mathrm{N} / \mathrm{C} .\) As a budding electrical engineer for Live- Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be and (b) find the maximum charge these plates can hold.

Two large metal parallel plates carry opposite charges of equal magnitude. They are separated by \(45.0 \mathrm{mm},\) and the potential difference between them is 360 \(\mathrm{V}\) (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 \(+2.40 \mathrm{nC}\) ?

(a) If an electron and a proton each have a kinetic energy of 1.00 eV, how fast is each one moving? (b) What would be their speeds if each had a kinetic energy of 1.00 \(\mathrm{keV}\) ? (c) If they were each traveling at 1.00\(\%\) the speed of light, what would be their kinetic energies in keV?

Axons. Neurons are the basic units of the nervous system. They contain long tubular structures called axons that propagate electrical signals away from the axon contains a solution axon contains a solution of potassium ions \(\mathrm{K}^{+}\) and large negative organic ions. The axon membrane prevents the large ions from leaking out, but the smaller \(\mathrm{K}^{+}\) ions are able to penetrate the membrane to some degree. (See Figure 18.39 . ) This leaves an excess negative charge on the inner surface of the axon membrane and an excess of positive charge on the outer surface, resulting in a potential difference across the membrane that prevents further \(\mathrm{K}^{+}\) ions from leaking out. Measurements show that this potential difference is typically about 70 \(\mathrm{mV}\) . The thickness of the axon membrane itself varies from about 5 to \(10 \mathrm{nm},\) so we'll use an average of 7.5 \(\mathrm{nm}\) . We can model the membrane as a large sheet having equal and opposite charge densities on its faces. (a) Find the electric field inside the axon membrane, assuming (not too realistically) that it is filled with air. Which way does it point, into or out of the axon? (b) Which is at a higher potential, the inside surface or the outside surface of the axon membrane?

A 5.80\(\mu\) F parallel-plate air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 \(\mathrm{V}\) . Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3} .\)

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