Problem 21 - University Physics with Modern Physics

Question

A positive point charge \(q\) is placed on the \(+y\) -axis at \(y=a\) and a negative point clarge \(-q\) is placed on the \(-y\) -axis at \(y=-a\) . A negative point charge \(-Q\) is located at some point on the \(+x\) -axis, (a) In a free- body diagram, show the forces that act on the charge \(-Q\) . (b) Find the \(x\) - and \(y\) -components of the net force that the two charges \(q\) and \(-q\) exert on \(-Q .\) (Your answer should involve only \(k, q, Q, a\) and the coordinate \(x\) of the third charge. \()\) (c) What is the net force on the charge \(-Q\) when it is at the origin \((x=0)\) ? (d) Graph the \(y\) -component of the net force on the charge \(-Q\) as a function of \(x\) for values of \(x\) between \(-4 a\) and \(+4 a\) .

Step-by-Step Solution

Answer

(a) Free-body diagram shows forces from \(+q\) and \(-q\). (b) Net force components: \(F_{net_x} = 0\), \(F_{net_y} = -2 \frac{k q Q a}{(x^2 + a^2)^{3/2}}\). (c) At origin, \(F_{net_y} = -2 \frac{k q Q}{a^2}\). (d) Graph \(F_{net_y}\) as symmetric, peaking near \(\pm a\).

Step 1: Understanding the Setup

We have three charges: a positive charge \(+q\) at \((0, a)\), a negative charge \(-q\) at \((0, -a)\), and a negative charge \(-Q\) placed at some point on the \(+x\)-axis, \((x, 0)\). We need to analyze the forces each charge exerts on \(-Q\).

Step 2: Identify Forces in Free-Body Diagram

Draw a free-body diagram for the charge \(-Q\). The positive charge \(+q\) exerts an attractive force directed towards \(+q\), and the negative charge \(-q\) exerts a repulsive force directed away from \(-q\). These forces will have both \(x\) and \(y\) components.

Step 3: Calculate Force from +q

The force exerted by \(+q\) on \(-Q\) is attractive. Using Coulomb's law, the magnitude of the force is \[ F_{+q} = \frac{k |q| |Q|}{(x^2 + a^2)}. \]The angle \(\theta\) between this force and the \(x\)-axis is given by \(\tan \theta = \frac{a}{x}\). So,\[ F_{+q_x} = F_{+q} \cdot \frac{x}{\sqrt{x^2+a^2}}, \quad F_{+q_y} = F_{+q} \cdot \frac{a}{\sqrt{x^2+a^2}}. \]

Step 4: Calculate Force from -q

The force exerted by \(-q\) on \(-Q\) is repulsive. Similarly, the magnitude of the force is \[ F_{-q} = \frac{k |q| |Q|}{(x^2 + a^2)}. \]By symmetry, the angle \(\theta\) is the same as for \(+q\). So,\[ F_{-q_x} = -F_{-q} \cdot \frac{x}{\sqrt{x^2+a^2}}, \quad F_{-q_y} = -F_{-q} \cdot \frac{a}{\sqrt{x^2+a^2}}. \]

Step 5: Find Net Force Components

The net force acting on \(-Q\) is the sum of the forces exerted by \(+q\) and \(-q\). So,\[ F_{net_x} = F_{+q_x} + F_{-q_x} = 0 \]because they are equal and opposite. For the \(y\)-component, we have\[ F_{net_y} = F_{+q_y} + F_{-q_y} = -2 \cdot \frac{k q Q a}{(x^2 + a^2)^{3/2}}. \]

Step 6: Net Force at the Origin (c)

When \(-Q\) is at the origin, \(x = 0\).The \(y\)-component of the net force is\[ F_{net_y} = -2 \cdot \frac{k q Q a}{a^3} = -2 \cdot \frac{k q Q}{a^2}. \]The \(x\)-component remains zero.

Step 7: Graphing the y-Component (d)

To graph the \(y\)-component of the force as a function of \(x\), plot \[ F_{net_y} = -2 \cdot \frac{k q Q a}{(x^2 + a^2)^{3/2}} \]for \(x\) values ranging from \(-4a\) to \(+4a\). This function is symmetric around the origin and shows maximum magnitude as \(x\) approaches \(±a\).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle that helps us understand how charges interact. It states that the force between two point charges is directly proportional to the product of their magnitudes, inversely proportional to the square of the distance between them, and acts along the line joining the charges. The formula is given by:\(\[ F = \frac{k |q_1| |q_2|}{r^2} \]\)where \(F\) is the force between the charges, \(q_1\) and \(q_2\) are the magnitudes of the charges, \(r\) is the distance between the charges, and \(k\) is Coulomb's constant (\(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\)).

• This law helps calculate both the attraction and repulsion forces based on charge properties.


• It applies to both static (charge at rest) and dynamic (charge in motion) electric field scenarios.

In problems involving multiple charges, such as in this exercise, Coulomb's Law is vital for determining the forces each charge exerts on another.

Free-Body Diagram

A free-body diagram is an essential tool in physics for visualizing the forces acting on an object. To construct a free-body diagram, follow these steps:

• Identify the object of interest, in this case, the charge $-Q$.


• Draw the object as a simple point or box.


• Represent each force acting on the object with arrows pointing in the direction of the force, labeled with force names.

In our exercise, the forces in play are:

• The attractive force from the positive charge $+q$ directed towards $+y$.


• The repulsive force from the negative charge $-q$ also directed towards $-y$.

The diagram serves as a clear visual representation, helping us to break down the forces into components and solve for net effects.

Point Charge

Point charges are an idealized model where the charge is assumed to be concentrated at a single point in space. This simplification helps in calculating electric forces and fields without worrying about the object's size or shape. There are some key aspects to consider:

• Point charges can be positive or negative, influencing whether they exert attractive or repulsive forces on other charges.


• They are assumed to have no dimensions, which simplifies the application of Coulomb's Law and diagrammatic representations.

In our given setup:

• The charge $+q$ is a point charge located at $(0, a)$, exerting an attractive force.


• The charge $-q$ is at $(0, -a)$, generating a repulsive force.

Understanding the characteristics of point charges is crucial for accurately analyzing and calculating forces in electric fields.

Net Force Calculation

Calculating the net force is the final step that sums up all the forces acting on a charge to determine its resultant motion or equilibrium.Firstly, we dissect the forces into their components:

• The \(x\)-components of the forces from \(+q\) and \(-q\) on \(-Q\) cancel each other out because they act in opposite directions and are equal in magnitude.
• For the \(y\)-components, both act downwards towards the origin, so they add up, leading to a net force in the \(y\) direction.

The net force can be calculated using:\[ F_{net_y} = -2 \cdot \frac{k q Q a}{(x^2 + a^2)^{3/2}} \] When \(-Q\) is placed at the origin \((x = 0)\): \[ F_{net_y} = -2 \cdot \frac{k q Q}{a^2} \]By understanding how to break forces into components and add them together, net force calculations become a straightforward process that reveals the overall effect of multiple forces on a point charge.