Question

An electric field is established in a nonuniform rod. A voltmeter is used to measure the potential difference between the left end of the rod and a point a distance \(x\) from the left end. The process is repeated, and it is found that the data are described by the relationship \(\Delta V=270 x^{2},\) where \(\Delta V\) has the units \(\mathrm{V} / \mathrm{m}^{2}\). What is the \(x\) -component of the electric field at a point \(13 \mathrm{~cm}\) from the left end?

Short answer

Answer: The x-component of the electric field at a point 13 cm from the left end is -7.02 × 10² V/m.

Step-by-step solution

Write down the given relationship

We are given the relationship between the potential difference \(\Delta V\) and the distance \(x\) as \(\Delta V = 270x^2\), where \(\Delta V\) has units of V/m².

Recall the relationship between electric field and potential

The relationship between the electric field \(E\) and the potential difference \(\Delta V\) is given by \(\Delta V = -\int E dx\), where the \(x\)-component of the electric field is \(E_x\).

Differentiate the potential difference with respect to x

To find the electric field at a specific point, we need to differentiate the potential difference with respect to distance \(x\). Let \(\frac{d(\Delta V)}{dx} = \frac{d(270x^2)}{dx}\). Differentiating \(270x^2\) with respect to \(x\), we get \(\frac{d(\Delta V)}{dx} = 540x\).

Find the electric field at the given point

Now, we can find the \(x\)-component of the electric field at the given point by using the result from step 3 as the \(x\)-component of the electric field is given by the negative of the derivative of potential difference with respect to x: \(E_x = -\frac{d(\Delta V)}{dx}\). At a point 13 cm from the left end, \(x = 13 \times 10^{-2}\) meters. Plug this value into our result from step 3: \(E_x = -540x = -540(13 \times 10^{-2}) = -7.02 \times 10^2 \mathrm{V/m}\) The \(x\)-component of the electric field at a point \(13 \mathrm{~cm}\) from the left end is \(-7.02 \times 10^2 \mathrm{V/m}\).

Key concepts

Potential Difference

When dealing with electric fields, the potential difference is a crucial concept. It refers to the difference in electric potential energy per unit charge between two points in space. In simpler terms, it tells us how much work is done by an electric field in moving a charge from one point to another.
In our exercise, the voltmeter measures the potential difference between the left end of the rod and a point at a certain distance, described by the equation \[\Delta V = 270 x^2\].
This equation indicates that the potential difference depends on the square of the distance from the starting point. It's important to note that potential difference is not always the same along the rod, especially when the electric field is nonuniform like in this scenario.

• Potential difference is key to understanding how energy varies in an electric field.
• It reflects the movement of charge and thus generates insight into the behavior of electric fields.
• Measured in volts, it acts as a bridge to other concepts like electric field strength.

Differentiation

Differentiation is a mathematical technique used to find the rate at which a quantity changes. In physics, it often helps in understanding how various parameters evolve over time or space. In our specific problem, differentiation provides us the tool to link the potential difference with the electric field.
The given potential difference \(\Delta V = 270x^2\) is differentiated with respect to distance \(x\) to establish a relationship with the electric field. This process is represented mathematically as:\[\frac{d(\Delta V)}{dx} = 540x\]This derivative tells us how the potential difference changes momentarily with a small change in \(x\). In simpler terms, it reveals the 'slope' of the potential difference at any given point, which is directly related to the electric field's strength and direction at that point.

• Differentiating processes help translate static equations into dynamic insights.
• It offers an immediate understanding of changes, crucial for electric field calculations.
• In electric fields, differentiation provides the fundamental step to finding field strength.

Electric Field Calculation

Calculating an electric field involves understanding its strength and how it affects charges within it. The electric field's \(x\)-component, \(E_x\), is found by taking the negative of the derivative of the potential difference, obtained through differentiation. The equation linking these is crucial:\[E_x = -\frac{d(\Delta V)}{dx}\]Using this approach, we calculated the electric field at a specific distance, \(x = 13 \text{ cm} = 13 \times 10^{-2} \text{ m}\), from the equation:\[E_x = -540(13 \times 10^{-2}) = -7.02 \times 10^2 \text{ V/m}\]This tells us that the electric field at this particular point is strong and points in the negative direction, signifying an attraction towards the negative side of the rod or a repulsion from the positive side.

• The electric field indicates the force per unit charge experienced by a charge in the field.
• Calculation involves both the magnitude and direction, informing us not just how hard, but also where the force acts.
• Strong electric fields suggest strong forces that move charges substantially within the field.