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A crude device for measuring charge consists of two small insulating spheres of radius \(a\), one of which is fixed in position. The other is movable along the \(x\) axis and is subject to a restraining force \(k x\), where \(k\) is a spring constant. The uncharged spheres are centered at \(x=0\) and \(x=d\), the latter fixed. If the spheres are given equal and opposite charges of \(Q / C\), obtain the expression by which \(Q\) may be found as a function of \(x\). Determine the maximum charge that can be measured in terms of \(\epsilon_{0}, k\), and \(d\), and state the separation of the spheres then. What happens if a larger charge is applied?

Short Answer

Expert verified
Answer: The expression for the charge Q in terms of the separation distance x is given by: \(Q(x) = \sqrt{kx(x+d)^2}\). When a larger charge than the maximum charge (\(Q_{max}\)) is applied to the spheres, the electric force between the spheres becomes too strong for the spring force to balance. This causes the movable sphere to continue moving away from the fixed sphere which can lead to inaccurate measurement of the charge and potentially damage or destroy the device.

Step by step solution

01

Analyzing the forces

There are two forces acting on the movable sphere: the electric force, which is repulsive, and the restraining force (spring force), which is attractive. We'll start by writing expressions for both of these forces. The electric force between two charged spheres with charges \(Q\) and \(-Q\) separated by a distance \(x+d\) can be given by Coulomb's Law: \(F_e = \frac{1}{4\pi \epsilon_0}\frac{Q^2}{(x+d)^2}\), where \(\epsilon_0\) is the vacuum permittivity. The restraining force or spring force on the sphere is given by Hooke's Law: \(F_s = kx\), where \(k\) is the spring constant.
02

Setting up the equilibrium

At the equilibrium position, the electric force and the spring force must balance each other, i.e., \(F_e = F_s\). Substituting the expressions from Step 1, we have: \(\frac{1}{4\pi \epsilon_0}\frac{Q^2}{(x+d)^2} = kx\)
03

Solving for Q

Now, we'll solve the equation above for the charge \(Q\). Before doing that, notice that multiplying both sides by \(4\pi \epsilon_0\) gives: \(Q^2 = kx (x+d)^2\). Taking the square root of both sides, we obtain the expression for the charge \(Q\) as a function of \(x\): \(Q(x) = \sqrt{kx(x+d)^2}\).
04

Finding the maximum charge (Q_max)

To determine the maximum charge that can be measured, we can differentiate the expression \(Q(x)\) with respect to \(x\) and set it to zero: \(\frac{dQ}{dx}(x) = 0\). Taking the derivative, we get: \(\frac{dQ}{dx} = \frac{1}{2} \frac{d}{dx}[kx(x+d)^2] \cdot (kx(x+d)^2)^{-\frac{1}{2}} = 0\). The derivative of \(kx(x+d)^2\) with respect to \(x\) is: \(\frac{d}{dx}[kx(x+d)^2] = k[(x+d)^2 + 2x(x+d)]\). Setting \(\frac{dQ}{dx} = 0\), we get: \(0 = k[(x+d)^2 + 2x(x+d)]\). This simplifies to: \(0 = (x+d) (x+3d)\). This gives us two possible values for \(x\), \(x = -d\) and \(x = -3d\). However, since the negative value of \(x\) doesn't make sense in this context, we'll choose \(x = -3d\). Therefore, the spheres are separated only by their radius, and we have the modified spring constant is \(\frac{k}{3}\) Now plugging this value of \(x\) into the expression for \(Q(x)\), we find the maximum charge: \(Q_{max} = \sqrt{k(-3d)(-2d)^2} = \sqrt{6kd^3}\).
05

What happens when a larger charge is applied?

If a larger charge than \(Q_{max}\) is applied to the spheres, the electric force between the spheres becomes too strong for the spring force to balance. As a result, the movable sphere will continue to move away from the fixed sphere, and the measurement of the charge will no longer be accurate. In the limiting case, the movable sphere would be torn from its anchor and forced away by the greater charge, leading to physical deformation or even destruction of the device.

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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 foundational principle describing the electrostatic force between two charged objects. It states that the magnitude of the force between two point charges is directly proportional to the product of the charges' magnitudes and inversely proportional to the square of the distance separating them. Mathematically, it is expressed as: \[ F_e = \frac{1}{4\pi \epsilon_0}\frac{Q^2}{(x+d)^2} \]Here, \( F_e \) is the electric force, \( Q \) is the magnitude of the charges, \( d \) is the distance apart before the moving sphere shifts, and \( x \) is the distance moved by the sphere along the axis. The constant \( \epsilon_0 \), known as the vacuum permittivity, is essential in quantifying how charges interact in a vacuum.
  • Direct proportionality: As the charge increases, the force increases.
  • Inverse square law: As the distance increases, the force decreases rapidly.
Coulomb's Law helps us understand that in this setup, the electrostatic force pushes the spheres apart as they carry opposite charges, setting the stage for other forces to counterbalance.
Spring Force
The spring force comes into play due to Hooke's Law. It is the force exerted by a spring that is compressed or stretched from its natural length. This physical setup uses a spring-like mechanism to measure the movement of the charging spheres. The spring force \( F_s \) is described by:\[ F_s = kx \]Here, \( k \) is the spring constant, representing the stiffness of the spring, and \( x \) is the displacement from the equilibrium position.
  • Restoring Force: It acts in the direction opposite to displacement, pushing the sphere back to the position.
  • Proportionality: The force increases linearly with the displacement \( x \).
In the given device, the spring force acts to draw the movable sphere back towards its starting position against the Coulomb's force, which helps in measuring the charge indirectly by balancing it at equilibrium.
Equilibrium
Equilibrium in this context refers to the point where the forces acting on the system are perfectly balanced. When the spring force equals the electrostatic force, the movable sphere stops moving, which defines the point of equilibrium.At equilibrium, the forces can be equated as follows:\[ \frac{1}{4\pi \epsilon_0}\frac{Q^2}{(x+d)^2} = kx \]Finding equilibrium helps in determining the value of the charge based on known constants and measured displacement. Balancing these forces reveals the conditions under which the movable sphere remains stationary, providing a mechanism for charge measurement.
  • Balance of Forces: No net force means stable equilibrium.
  • Measurement Point: At equilibrium, one can solve for \( Q(x) \).
Maximum Charge Measurement
To find the maximum charge that the crude device can measure, we optimize for \( Q(x) \) by finding the point at which its derivative equals zero. This involves setting the modified expression, derived from equating forces, to find the critical points and evaluating the situation where the spring force can no longer contain the expanding electrostatic force.We calculate:\[Q(x) = \sqrt{kx(x+d)^2}\]Differentiating and setting the derivative to zero gives the maximum, leading to:\[Q_{max} = \sqrt{6kd^3}\]Here, \( d \) effectively plays a key role in limiting movement. If the charge applied to each sphere exceeds \( Q_{max} \), the electrical force overpowers the restraining ability of the spring, potentially causing the movable sphere to break free or disrupt measurement accuracy.

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Most popular questions from this chapter

A uniform line charge of \(2 \mu \mathrm{C} / \mathrm{m}\) is located on the \(z\) axis. Find \(\mathbf{E}\) in rectangular coordinates at \(P(1,2,3)\) if the charge exists from \((a)-\infty<\) \(z<\infty ;(b)-4 \leq z \leq 4\).

If \(\mathbf{E}=20 e^{-5 y}\left(\cos 5 x \mathbf{a}_{x}-\sin 5 x \mathbf{a}_{y}\right)\), find \((a)|\mathbf{E}|\) at \(P(\pi / 6,0.1,2) ;(b)\) a unit vector in the direction of \(\mathbf{E}\) at \(P ;(c)\) the equation of the direction line passing through \(P\).

Find \(\mathbf{E}\) at the origin if the following charge distributions are present in free space: point charge, \(12 \mathrm{nC}\), at \(P(2,0,6) ;\) uniform line charge density, \(3 \mathrm{nC} / \mathrm{m}\), at \(x=-2, y=3 ;\) uniform surface charge density, \(0.2 \mathrm{nC} / \mathrm{m}^{2}\) at \(x=2\).

The electron beam in a certain cathode ray tube possesses cylindrical symmetry, and the charge density is represented by \(\rho_{v}=-0.1 /\left(\rho^{2}+10^{-8}\right)\) \(\mathrm{pC} / \mathrm{m}^{3}\) for \(0<\rho<3 \times 10^{-4} \mathrm{~m}\), and \(\rho_{v}=0\) for \(\rho>3 \times 10^{-4} \mathrm{~m} .(a)\) Find the total charge per meter along the length of the beam; \((b)\) if the electron velocity is \(5 \times 10^{7} \mathrm{~m} / \mathrm{s}\), and with one ampere defined as \(1 \mathrm{C} / \mathrm{s}\), find the beam current.

Eight identical point charges of \(Q \mathrm{C}\) each are located at the corners of a cube of side length \(a\), with one charge at the origin, and with the three nearest charges at \((a, 0,0),(0, a, 0)\), and \((0,0, a)\). Find an expression for the total vector force on the charge at \(P(a, a, a)\), assuming free space.

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