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For each of the following arrangements of two point charges, find all the points along the line passing through both charges for which the electric potential \(V\) is zero (take \(V=0\) infinitely far from the charges) and for which the electric field \(E\) is zero: (a) charges \(+Q\) and \(+2 Q\) separated by a distance \(d\) , and (b) charges \(-Q\) and \(+2 Q\) separated by a distance \(d\) . (c) Are both \(V\) and \(E\) zero at the same places? Explain.

Short Answer

Expert verified
In both arrangements, zero points for \(V\) and \(E\) are not at the same places.

Step by step solution

01

Determine Electric Potential Formula

The electric potential at a point due to a point charge is given by the formula \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant and \( r \) is the distance from the charge. For multiple charges, the total potential is the algebraic sum of the potentials from each charge.
02

Find Zero Potential Points for (a) +Q and +2Q

For two charges \(+Q\) at \(x=0\) and \(+2Q\) at \(x=d\), the potential at a point \(x\) along the line is \( V = \frac{kQ}{x} + \frac{k(2Q)}{d-x} \). Set \( V = 0 \) and solve for \(x\). This simplifies to \( \frac{1}{x} + \frac{2}{d-x} = 0 \). This gives the condition \(2x = d-x\), yielding \( x = \frac{d}{3}\). The only physically accessible point between the charges is at infinity.
03

Determine Zero Electric Field Points for (a) +Q and +2Q

The electric field \(E\) from a point charge is \( E = \frac{kQ}{r^2} \). Since both fields are in the same direction (repulsive for positive charges), set the sum of the fields to zero: \( \frac{kQ}{x^2} = \frac{k(2Q)}{(d-x)^2} \). Solving gives \( x = \frac{d}{\sqrt{2} + 1} \). Physically, the zero electric field is located on the side of the charge +Q, external to the line connecting the charges.
04

Find Zero Potential Points for (b) -Q and +2Q

For charges \(-Q\) at \(x=0\) and \(+2Q\) at \(x=d\), the potential is \( V = \frac{-kQ}{x} + \frac{k(2Q)}{d-x} \). Set \( V = 0 \) and solve for \(x\). The equation leads to \( \frac{1}{x} = \frac{2}{d-x} \). Simplifying yields \( x = \frac{d}{3} \). Thus, a zero potential point is located at a position \( \frac{d}{3} \) from the negative charge towards the positive charge.
05

Determine Zero Electric Field Points for (b) -Q and +2Q

For the electric field with opposing directions for opposite charges, set \( \frac{k(-Q)}{x^2} + \frac{k(2Q)}{(d-x)^2} = 0 \). This results in \( x = \frac{d}{\sqrt{2} + 1} \). The zero electric field point lies beyond the negative charge \(-Q\), outside the line segment connecting the charges.
06

Check If Both V and E Are Zero at the Same Places

In both cases (a) and (b), the positions where the electric potential \( V \) is zero are different from where the electric field \( E \) is zero. Therefore, \( V \) and \( E \) are not zero at the same locations.

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Key Concepts

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

Point Charges
Point charges are fundamental concepts in electrostatics. These are hypothetical charges that are infinitely small, allowing us to treat them as having no physical size or shape. They only have a magnitude and sign, which can be positive or negative. These charges exert forces on each other, and the strength of these forces depends on the magnitudes of the charges and the distance between them.

In an arrangement involving multiple point charges, each charge contributes to the total electric potential and electric field at a given point in space. The interactions are governed by specific laws derived from experimental observations, crucial among them being Coulomb's Law, which describes the force between two point charges.

When analyzing problems with point charges, it's essential to understand how individual contributions combine. The electric potential at a point due to multiple charges is the algebraic sum of potentials from each charge, while the electric field is a vector sum of fields due to each charge. Therefore, both the magnitude and direction are important for electric fields, while only the magnitude matters for electric potentials.
Zero Potential Points
Electric potential is a scalar quantity, meaning it only has magnitude, not direction. The principle of superposition is important here, which states the total potential at any point is the sum of the potentials due to individual charges. A point of zero potential doesn't mean that there are no charges; it merely means that the algebraic sum of potentials at that point equals zero.

Consider two charges such as "+Q" and "+2Q." If you're looking for a zero potential point along the line connecting these charges, you need to find where their potential contributions cancel each other out. Using the formula for electric potential, set the total potential to zero and solve for the position. For instance, in the case of "+Q" at x=0 and "+2Q" at x=d, a point where the potential is zero will not lie between them, as both potentials have the same sign. Instead, zero potential points can be found conceptually by solving the equation that balances out the contributions.

For charges like "-Q" and "+2Q," the signs are different, allowing for a point between them where the potential can be zero, theoretically at one-third the separation distance from the negative charge.
Zero Electric Field Points
Electric fields are vector quantities, having both magnitude and direction. A point of zero electric field is a point where the net electric field due to all charges is zero. Unlike potential, it implies that the forces from the fields cancel each other out, including their directional components.

To find zero electric field points, you must consider the vector nature of the fields. For example, between two charges like "+Q" and "+2Q," the fields will point away from each charge, and a zero electric field point can exist outside their line if their field magnitudes cancel each other out. This location is calculated by setting the field magnitudes equal and solving for the position.

In configurations like "-Q" and "+2Q," the fields point in opposite directions between the charges, offering a potential zero point within the region between the charges. Solving for these positions relies on balancing their contributions respective to their vector directions.
Coulomb's Law
Coulomb's Law is fundamental to understanding electric forces between point charges. It states that the force between two point charges varies directly with the product of the magnitudes of the charges and inversely with the square of the distance separating them. Mathematically, we express this as \[F = \frac{k |Q_1 Q_2|}{r^2}\]where:- \( F \) is the magnitude of the force between the charges.- \( Q_1 \) and \( Q_2 \) are the point charges.- \( r \) is the distance between the charges.- \( k \) is the Coulomb's constant, approximately equal to \( 8.99 × 10^9 \, N \, m^2/C^2 \).

Coulomb's Law helps explain the behavior of charged particles and is essential when calculating forces in electrostatic scenarios. It emphasizes how charges of the same sign repel each other, while opposite charges attract. Notably, Coulomb's law is a key element in determining individual electric fields, a crucial step in finding zero potential and zero electric field points. By understanding the interplay of these forces, we can accurately predict how point charges will interact in various configurations.

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Most popular questions from this chapter

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?

Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1000 \(\mathrm{km} / \mathrm{s}\) , measured relative to the earth. Find the maximum electrical force that these protons will exert on each other.

A very long cylinder of radius 2.00 \(\mathrm{cm}\) carries a uniform charge density of 1.50 \(\mathrm{nC} / \mathrm{m}\) . (a) Describe the shape of the equipotential surfaces for this cylinder. (b) Taking the reference level for the zero of potential to be the surface of the cylinder, find the radius of equipotential surfaces having potentials of \(10.0 \mathrm{V}, 20.0 \mathrm{V},\) and 30.0 \(\mathrm{V}\) . (c) Are the equipotential surfaces equally spaced? If not, do they get closer together or farther apart as \(r\) increases?

An infinitely long line of charge has linear charge density \(5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton (mass \(1.67 \times 10^{-27} \mathrm{kg}\) . charge \(+1.60 \times 10^{-19} \mathrm{C} )\) is 18.0 \(\mathrm{cm}\) from the line and moving directly toward the line at \(1.50 \times 10^{3} \mathrm{m} / \mathrm{s}\) (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge? (Hints See Example \(23.10 . )\)

(a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00\(\mu \mathrm{C}\) and the other a charge of \(-3.50 \mu \mathrm{C},\) with their centers separated by a distance of 0.250 \(\mathrm{m}\) . Assume zero potential energy when the charges are infinitely separated. (b) Suppose that one of the spheres is held in place and the other sphere, which has a mass of 1.50 \(\mathrm{g}\) , is shot away from it. What minimum initial speed would the moving sphere need in order to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it was infinitely distant from the fixed sphere.)

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