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Chapter 23: Problem 77

Two fixed point charges are on the \(x\) -axis. A charge of \(-3.00 \mathrm{mC}\) is located at \(x=+2.00 \mathrm{~m}\) and a charge of \(+5.00 \mathrm{mC}\) is located at \(x=-4.00 \mathrm{~m}\) a) Find the electric potential, \(V(x),\) for an arbitrary point on the \(x\) -axis. b) At what position(s) on the \(x\) -axis is \(V(x)=0 ?\) c) Find \(E(x)\) for an arbitrary point on the \(x\) -axis.

Short Answer

Expert verified
Question: Determine the position(s) on the x-axis where the electric potential V(x) is equal to 0 given that the electric potential due to the first charge is \(V_1(x) = \dfrac{k(-3.00 * 10^{-3} C)}{x - 2}\) and the electric potential due to the second charge is \(V_2(x) = \dfrac{k(5.00 * 10^{-3} C)}{x + 4}\). Answer: To find the position(s) on the x-axis where V(x) is equal to 0, set the total electric potential equal to 0 and solve for x: \(0 = V_1(x) + V_2(x)\) Rearrange the equation: \(V_1(x) = -V_2(x)\) Then, solve for x to find the position(s) where the electric potential is 0.

Step by step solution

01

Find the electric potential due to each charge

To find the electric potential at any point on the x-axis, we first need to find the electric potential due to each point charge. For an arbitrary point x along the x-axis, the distance from the first charge to x is \(x - 2\) meters and the distance from the second charge to x is \(x + 4\) meters. The electric potential, V(x), due to a point charge q at a distance r, is given by: \(V(x) = \dfrac{kq}{r}\), where \(k = 8.99 * 10^9 N m^2/C^2\) is the Coulomb's constant. Thus, the electric potential due to each charge at an arbitrary point x can be found using the above formula: \(V_1(x) = \dfrac{k(-3.00 * 10^{-3} C)}{x - 2}\) (due to the first charge) \(V_2(x) = \dfrac{k(5.00 * 10^{-3} C)}{x + 4}\) (due to the second charge)
02

Calculate the total electric potential at any point along the x-axis

Now, to obtain the total electric potential (V(x)) at an arbitrary point x along the x-axis, we simply add the individual electric potentials due to each charge: \(V(x) = V_1(x) + V_2(x)\)
03

Determine the position(s) on the x-axis where V(x) is equal to 0

To find the position(s) along the x-axis where the electric potential is 0, we place the total electric potential V(x) equal to 0 and solve for x: \(0 = V_1(x) + V_2(x)\) Rearrange the equation: \(V_1(x) = -V_2(x)\) Then, solve for x to find the position(s) where the electric potential is 0.
04

Find the electric field at any arbitrary point x along the x-axis

To find the electric field (E(x)) at any point x along the x-axis due to these point charges, we use the equation for the electric field due to a point charge: \(E(x) = \dfrac{kq}{r^2}\) Hence, the electric field due to each charge at an arbitrary point x is given by: \(E_1(x) = - \dfrac{k(-3.00 * 10^{-3} C)}{(x - 2)^2}\) (due to the first charge) \(E_2(x) = \dfrac{k(5.00 * 10^{-3} C)}{(x + 4)^2}\) (due to the second charge)
05

Calculate the total electric field at any point along the x-axis

Now to calculate the total electric field (E(x)) at an arbitrary point x along the x-axis, we must add the individual electric fields due to each charge. However, since we are only concerned with the x-component of the electric field for this problem, we sum the x-components of the electric fields: \(E(x) = E_1(x) + E_2(x)\)

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle used to calculate the electric force between two point charges. It states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This can be expressed mathematically as:\[ F = \dfrac{k |q_1 q_2|}{r^2} \]Here, \(F\) is the magnitude of the force, \(q_1\) and \(q_2\) are the charges, \(r\) is the distance between the charges, and \(k = 8.99 \times 10^9 \ \text{Nm}^2/\text{C}^2\) is Coulomb's constant.
Coulomb's Law is similar to Newton's Law of Universal Gravitation but applies specifically to electric charges rather than masses.
  • It helps in understanding how forces act at a distance without any direct contact.
  • It is the basis for deriving other concepts like the electric field and electric potential.
Understanding this law is crucial in calculating the effects of electrostatic forces in various situations, including those involving multiple charges.
Electric Fields
An electric field is a vector field that shows the influence a charge exerts on other charges in the space around it. The strength and direction of an electric field at a point are defined as the force per unit positive charge at that point.
The formula to find the electric field \(E\) due to a point charge \(q\) is:\[ E = \dfrac{kq}{r^2} \]where \(r\) is the distance from the charge to the point where the field is being calculated, and \(k\) is Coulomb鈥檚 constant.
  • The direction of the electric field is the direction of the force it would exert on a positive test charge.
  • Fields are always directed away from positive charges and towards negative charges.
  • Electric field lines never intersect and are closer together where the field is stronger.
To calculate net electric fields in scenarios with multiple charges, each point charge鈥檚 field is calculated individually and then vectorally added based on direction and magnitude. Electric fields can tell you a lot about how charges will move in a particular setup.
Point Charges
Point charges are idealized as charges that are so small compared to the distances involved in the problem that they can be treated as if they are located at a single point in space.
Point charges are often used in theoretical problems to simplify calculations related to electrostatic force, electric fields, and electric potential.
  • They help to model real charges and provide insights into charge interactions without the complexity of actual charge distributions.
  • Using the concept of point charges, it becomes easier to apply laws like Coulomb's Law to calculate forces, potentials, and fields.
  • In calculations, point charges are typically defined with magnitude (in coulombs) and position.
Point charges represent an essential concept in physics, enabling the study of charge interactions in a straightforward manner.

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Most popular questions from this chapter

Derive an expression for electric potential along the axis (the \(x\) -axis) of a disk with a hole in the center, as shown in the figure, where \(R_{1}\) and \(R_{2}\) are the inner and outer radii of the disk. What would the potential be if \(R_{1}=0 ?\)

Why is it important, when soldering connectors onto a piece of electronic circuitry, to leave no pointy protrusions from the solder joints?

A hollow spherical conductor with a \(5.0-\mathrm{cm}\) radius has a surface charge of \(8.0 \mathrm{nC}\). a) What is the potential \(8.0 \mathrm{~cm}\) from the center of the sphere? b) What is the potential \(3.0 \mathrm{~cm}\) from the center of the sphere? c) What is the potential at the center of the sphere?

What potential difference is needed to give an alpha particle (composed of 2 protons and 2 neutrons) \(200 \mathrm{keV}\) of kinetic energy?

A point charge \(Q\) is placed a distance \(R\) from the center of a conducting sphere of radius \(a,\) with \(R>a\) (the point charge is outside the sphere). The sphere is grounded, that is, connected to a distant, unlimited source and/or sink of charge at zero potential. (Neither the distant ground nor the connection directly affects the electric field in the vicinity of the charge and sphere.) As a result, the sphere acquires a charge opposite in sign to \(Q\), and the point charge experiences an attractive force toward the sphere. a) Remarkably, the electric field outside the sphere is the same as would be produced by the point charge \(Q\) plus an imaginary mirror-image point charge \(q\), with magnitude and location that make the set of points corresponding to the surface of the sphere an equipotential of potential zero. That is, the imaginary point charge produces the same field contribution outside the sphere as the actual surface charge on the sphere. Calculate the value and location of \(q\). (Hint: By symmetry, \(q\) must lie somewhere on the axis that passes through the center of the sphere and the location of \(Q .)\) b) Calculate the force exerted on point charge \(Q\) and directed toward the sphere, in terms of the original quantities \(Q, R,\) and \(a\) c) Determine the actual nonuniform surface charge distribution on the conducting sphere.
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