Problem 22 - University Physics with Modern Physics

Question

Two positive point charges \(q\) are placed on the \(y\) -axis at \(y=a\) and \(y=-a .\) 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 positive charges exert on \(-Q\) . (Your answer \(r\) . 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 \((x=0) ?(\mathrm{d})\) Graph the \(x\) -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

The net force on \\(-Q\\) at the origin is zero. The \\x\\-component depends symmetrically on \\x\\.

Step 1: Analyze Forces on Charge -Q

The negative charge \(-Q\) experiences forces due to both positive charges \(+q\). Since these charges are symmetrically placed on the \(+y\) and \(-y\) axes, forces will be exerted along lines connecting the charges with \(-Q\). For both charges, the force will have a component along the \(-x\)-axis and opposite \(+y\) or \(-y\)-axes, depending on their position.

Step 2: Calculate Distance (r)

For each positive charge \(+q\), the distance to the negative charge \(-Q\) is calculated. By Pythagoras' theorem, the distance \r\ between \(-Q\) at \(x, 0\) and \(+q\) at \(0, \pm a\) is: \[r = \sqrt{x^2 + a^2}.\]

Step 3: Determine Force from a Single Charge (F)

Using Coulomb's Law, compute the force exerted by one \(+q\) charge on \(-Q\): \[F = \frac{kqQ}{r^2} = \frac{kqQ}{x^2 + a^2}.\]

Step 4: Find x-Component of Force (F_x)

Using trigonometry, find the \x\-component of this force from symmetry: \\[F_{x} = F \cdot \frac{x}{r} = \frac{kqQ \cdot x}{(x^2 + a^2)^{3/2}}.\] This is the same for both charges due to symmetry.

Step 5: Find y-Component of Force (F_y)

Similarly, the \y\-component of force \F_{y}\ for one charge is: \[F_{y} = F \cdot \frac{a}{r} = \frac{kqQ \cdot a}{(x^2 + a^2)^{3/2}}.\] Each exerts opposite and equal magnitudes, thus canceling out net \y\-components.

Step 6: Calculate Net Force at Origin (x=0)

At \(x=0\), \\[F_{net} = 2 \times \frac{kqQ \cdot 0}{a^3} = 0.\] The total force is zero due to symmetry.

Step 7: Graph x-Component of Net Force

Plot \F_x\ as a function of \x\ ranging from \-4a\ to \+4a\. The formula for \F_x\ is: \[F_x = 2 \times \frac{kqQ \cdot x}{(x^2 + a^2)^{3/2}}.\] Evaluate and draw the function, ensuring both side symmetries and zero crossing at \(x=0\).

Key Concepts

Coulomb's Law

Did you know that electricity can exert a force without touching? This phenomenon is known as electrostatic force. And it's all explained by Coulomb's Law. In simple terms, Coulomb's Law tells us how to calculate the force between two point charges.
The formula for this is: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \] Where:

• \( F \) is the magnitude of the force between the charges,
• \( k \) is Coulomb's constant, approximately equal to \( 8.99 \times 10^9 \) \( N \cdot m^2/C^2 \),
• \( q_1 \) and \( q_2 \) are the amounts of each charge,
• \( r \) is the distance between the centers of the two charges.

According to this law, like charges repel and opposite charges attract. So, if one charge is positive and the other is negative, they will pull toward each other. If both are negative or both are positive, they push away. This principle is crucial for predicting how charged particles interact.

Point Charges

In the world of electrostatics, we often talk about point charges. These are idealized charges that are considered to be concentrated at a single point in space.
Because point charges are modeled as having no size, they are extremely useful for simplifying calculations related to electric forces.
How do we use point charges in electrostatic problems? Let's break it down.

• We assume the charge is perfectly spherical and infinitesimally small.
• We can then treat it as a point source of the electric field.
• This allows us to apply Coulomb's Law effectively.

For example, in our exercise, the charges are positioned on the \( y \)-axis, and we treat them as point charges. This way, we can use symmetry and the distances calculated from their positions to find forces acting in the system. The simplicity of point charges makes complex problems much more approachable.

Net Force Calculation

When multiple forces act on an object, like our negative charge \(-Q\), we need to determine the net force to understand the overall effect. This involves calculating both the magnitude and direction of all individual forces and then summing them up vectorially.
In this case:

• The horizontal or \( x \)-component of the force from each positive charge is calculated using \[ F_{x} = \frac{kqQ \cdot x}{(x^2 + a^2)^{3/2}} \] and because both \( +q \) charges pull with equal and opposite horizontal components, the net force is simply their sum.
• For the vertical \( y \)-components, they are opposed and equal. Hence they cancel each other out, making net \( F_y \) zero.

Summing these vectors gives you the net force experienced by \(-Q\), showing that careful analysis of each component is necessary. At the origin, symmetry ensures the forces from both positive charges cancel perfectly, resulting in no net force.

Symmetry in Physics

Symmetry can make solving physics problems easier. It helps us predict and calculate forces and other attributes.
In the given exercise, both positive charges are placed symmetrically around the origin on the \( y \)-axis.
Let's see how symmetry aids our solution.

• Because the forces acting on \(-Q\) from both positive charges are equidistant and in opposite directions, they create equal but opposite vertical force components.
• This symmetry ensures that these vertical components cancel each other out, leaving no net force in the \( y \)-direction.
• In the horizontal direction, since both forces pull in the \(-x\)-direction, they add up rather than cancel out.

Utilizing symmetry allowed us to simplify the problem significantly. We derived that the net force would always be along the \( x \)-axis except when the negative charge is at the origin, further simplifying computations.