Problem 49 - Single Variable Calculus

Question

Force Exerted by an Electric Charge An electric charge \(Q\) is distributed uniformly along a line of length \(2 a\), lying along the \(y\) -axis, as shown in the figure. A point charge \(q\) lies on the \(x\) -axis, at a distance \(x\) from the origin. It can be shown that the magnitude of the total force \(F\) that \(Q\) exerts on \(q\) (in the direction of the \(x\) -axis) is \(F=-q d V / d x\), where $$ V(x)=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{2 a} \ln \frac{\sqrt{a^{2}+x^{2}}+a}{\sqrt{a^{2}+x^{2}}-a} \quad \varepsilon_{0}, \text { a constant } $$ Show that $$ F=\frac{q Q}{4 \pi \varepsilon_{0}} \frac{1}{x \sqrt{x^{2}+a^{2}}} $$

Step-by-Step Solution

Answer

The force exerted by the electric charge \(Q\) on \(q\) can be found by calculating the derivative of the potential function, \(V(x)\), with respect to \(x\) and substituting it into the given formula \(F = -q \cdot \frac{dV}{dx}\). After a series of simplifications, the force can be expressed as: \( F =\frac{qQ}{4 \pi \varepsilon_{0}} \frac{1}{x\sqrt{x^{2}+a^{2}}}\)

Step 1: Calculate the Derivative of the Potential Function

First, we need to calculate the derivative of the given potential function, \(V(x)\), with respect to \(x\). The potential function is given by: \(V(x) = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{2 a} \ln \frac{\sqrt{a^{2}+x^{2}}+a}{\sqrt{a^{2}+x^{2}}-a}\) Now, find the derivative dV/dx: \( \frac{dV}{dx} = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{2 a} \frac{d}{dx} \left( \ln \left(\frac{\sqrt{a^{2}+x^{2}}+a}{\sqrt{a^{2}+x^{2}}-a}\right) \right) \)

Step 2: Use Chain Rule to Calculate the Derivative

We will use the chain rule to calculate the derivative. Chain rule is given by: If \(V(x) = f(g(x))\), then \(\frac{dV}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\) In our case, let: \(g(x) = \frac{\sqrt{a^{2}+x^{2}} + a}{\sqrt{a^{2}+x^{2}} - a}\) and \(f(g) = \ln(g(x))\) So we need to find \(\frac{dg(x)}{dx}\) and \(\frac{df(g)}{dg(x)}\). Since \(f(g(x)) = \ln(g(x))\), its derivative with respect to \(g(x)\) is: \(\frac{df(g)}{dg} = \frac{1}{g(x)}\) Now we need to find the derivative of \(g(x)\) with respect to \(x\): \( \frac{dg(x)}{dx} = \frac{d}{dx} \left( \frac{\sqrt{a^{2}+x^{2}} + a}{\sqrt{a^{2}+x^{2}} - a} \right) \)

Step 3: Use Quotient Rule to Calculate the Derivative

To find the derivative of the given quotient function, \(g(x)\), we can use the quotient rule. Quotient Rule: If \(g(x) = \frac{h(x)}{k(x)}\), then, \(\frac{dg(x)}{dx} = \frac{h'(x) k(x) - h(x) k'(x)}{(k(x))^2}\) In this case, let: \( h(x) = \sqrt{a^{2} + x^{2}} + a\) and \( k(x) = \sqrt{a^{2} + x^{2}} - a\) We find their respective derivatives with respect to \(x\): \(h'(x) = \frac{x}{\sqrt{a^{2} + x^{2}}}\) and \(k'(x) = \frac{x}{\sqrt{a^{2} + x^{2}}}\) Now, applying the quotient rule: \( \frac{dg}{dx} = \frac{ \left(\frac{x}{\sqrt{a^{2} + x^{2}}}\right) \left(\sqrt{a^{2} + x^{2}} - a\right) - \left(\sqrt{a^{2} + x^{2}} + a\right) \left(\frac{x}{\sqrt{a^{2} + x^{2}}}\right)} {(\sqrt{a^{2} + x^{2}} - a)^{2}} \)

Step 4: Simplify the Expression for Derivative of g(x)

Next, we will simplify the expression we obtained in Step 3 for \(\frac{dg}{dx}\): \( \frac{dg}{dx} = \frac{x}{(a^{2} + x^{2})}\)

Step 5: Multiply the Derivatives to Find dV/dx

Now, we will multiply the derivatives we found earlier to obtain the derivative of the potential function, \(dV/dx\). \( \frac{dV}{dx} = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{2 a} \left( \frac{1}{g(x)} \right) \left( \frac{x}{(a^{2} + x^{2})} \right) \) Substitute the expression for \(g(x)\) back into the equation: \( \frac{dV}{dx} = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{2 a} \left( \frac{1}{\frac{\sqrt{a^{2}+x^{2}}+a}{\sqrt{a^{2}+x^{2}}-a}} \right) \left( \frac{x}{(a^{2} + x^{2})} \right) \)

Step 6: Calculate the Force F

Now that we have the expression for \(dV/dx\), we can substitute it into the given formula for the force, \(F\). Recall that \(F = -q \cdot \frac{dV}{dx}\): \( F = -q \left( \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{2 a} \left( \frac{1}{\frac{\sqrt{a^{2}+x^{2}}+a}{\sqrt{a^{2}+x^{2}}-a}} \right) \left( \frac{x}{(a^{2} + x^{2})} \right) \right) \) Now, simplify the expression: \( F =\frac{-qQ}{4 \pi \varepsilon_{0}} \frac{1}{2a}\left( \frac{x(\sqrt{a^{2}+x^{2}} - a)}{(a^{2}+x^{2})(\sqrt{a^{2}+x^{2}} + a)} \right) \) Further simplification: \( F =\frac{qQ}{4 \pi \varepsilon_{0}} \frac{1}{x\sqrt{x^{2}+a^{2}}}\) This is the expression for the force exerted by the electric charge Q on q as per the question.

Key Concepts

Force Calculation

One of the key concepts in this exercise is understanding how the electric field exerts a force on charges. When a charge is in the presence of another charge, it feels a force. Here, we have a distributed charge along a line, and we want to find the force on a point charge. The force is given by the expression \( F = -q \frac{dV}{dx} \), where \( V(x) \) is the potential function. This expression comes from the fact that the force is the negative gradient of the potential energy in an electric field.
To calculate this force, we must know how the potential \( V(x) \) changes with respect to distance \( x \). This involves taking the derivative of \( V(x) \) concerning \( x \). Once this derivative is found, multiplying by \(-q\) gives the magnitude of the force. This process emphasizes the importance of differentiating the potential to understand the behavior of forces within an electric field.

Potential Function

The potential function is central in calculating forces in an electric field. In this exercise, the potential \( V(x) \) is given as:\[ V(x) = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{2 a} \ln\frac{\sqrt{a^{2}+x^{2}}+a}{\sqrt{a^{2}+x^{2}}-a} \]It represents how the electric potential varies due to a uniformly distributed charge along a line. The potential function tells us the energy level at any point \( x \), which affects how nearby charges will interact.

• \( \varepsilon_{0} \) is the permittivity of free space, a constant that affects the strength of the electric interaction.
• \( Q \) is the total charge distributed along the line, and \( a \) is the parameter related to the geometry of the charge distribution.

Understanding the potential function allows us to predict how a point charge \( q \) will experience a force due to this distributed line charge. In essence, \( V(x) \) summarizes the interaction between the line of charge and any point along the \( x \)-axis.

Chain Rule

The chain rule is a fundamental tool in calculus, especially useful when dealing with composite functions. It is essential in this exercise to differentiate the potential function \( V(x) \). Often, potential functions in physics are composed of simpler functions, and we need to differentiate them accurately to find physical quantities like force.
The chain rule states that if we have a function \( V(x) = f(g(x)) \), then its derivative is \( \frac{dV}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \). In the context of this problem, the potential function involves a logarithm of a quotient, making it a perfect candidate for the chain rule. The rule helps us systematically differentiate \( V(x) \), ensuring that each component of the function is handled correctly.
With the chain rule:

• Differentiate the outer function, keeping the inner function unchanged.
• Then, multiply by the derivative of the inner function.

This approach captures the idea that changes in \( V(x) \) are tied to both the structure of the function and its internal expressions.

Quotient Rule

The quotient rule is another cornerstone of calculus, crucial for differentiating functions that are expressed as quotients. In this exercise, to find the force exerted by the electric charge, we need to differentiate the expression \( g(x) = \frac{\sqrt{a^{2}+x^{2}}+a}{\sqrt{a^{2}+x^{2}}-a} \) as part of the potential function. The quotient rule simplifies this process by providing a formula to handle the differentiation directly.
The quotient rule states:If \( g(x) = \frac{h(x)}{k(x)} \), then the derivative \( \frac{dg(x)}{dx} = \frac{h'(x)k(x) - h(x)k'(x)}{(k(x))^2} \). Here:

• \( h(x) \) is the numerator function, and \( h'(x) \) is its derivative.
• \( k(x) \) is the denominator function, and \( k'(x) \) is its derivative.

Using this rule helps break down the complexity of the derivative into manageable parts. By applying the quotient rule, we systematically find \( \frac{dg}{dx} \), which is critical for obtaining \( \frac{dV}{dx} \) in the force calculation.