Question

A thin, flat washer is a disk with an outer diameter of \(10.0 \mathrm{~cm}\) and a hole of diameter \(4.00 \mathrm{~cm}\) in the center. The washer has a uniform charge distribution and a total charge of \(7.00 \mathrm{nC}\). What is the electric field on the axis of the washer at a distance of \(30.0 \mathrm{~cm}\) from the center of the washer?

Short answer

Answer: The electric field on the axis of the washer at a distance of 30.0 cm from the center is approximately 1542.5 N/C.

Step-by-step solution

Expression for Electric Field Due to an Infinitely Thin Ring of Charge

The electric field due to an infinitely thin ring of charge, with a radius r and an infinitesimal thickness dr, can be written as: dE = \frac{k dq}{(x^2 + r^2)^{3/2}} \cdot x where - k is the electrostatic constant, \(k = 8.99 \times 10^9 \mathrm{~N m^2 / C^2}\) - dq is the infinitesimal charge on the ring, which can be written as dq = σ * 2πr dr (σ is the surface charge density) - x is the distance from the center of the washer along the axis (given as 30.0 cm)

Integrate to Sum Up the Electric Fields due to the Rings

Next, we will integrate the expression from Step 1 to sum up the electric fields due to each infinitesimally thin ring on the washer: E = ∫(dE) = ∫\frac{k dq}{(x^2 + r^2)^{3/2}} \cdot x Since the total charge on the washer (7.00 nC) is uniformly distributed, we can write the surface charge density σ as: σ = \frac{Q_{total}}{Area} = \frac{7.00 \times 10^{-9} C}{\pi((0.1)^2 - (0.04)^2) m^2} Now, plug σ into the expression for dq (dq = σ * 2πr dr), and substitute dq into the expression for E: E = ∫\frac{k (σ * 2πr dr)}{(x^2 + r^2)^{3/2}} \cdot x The limits of integration for the radius r will be from the inner diameter (2.00 cm) to the outer diameter (5.00 cm). Thus, we will integrate with respect to r from 0.02 m to 0.05 m: E = ∫_{0.02}^{0.05}\frac{k (σ * 2πr dr)}{(x^2 + r^2)^{3/2}} \cdot x

Evaluate the Integral

The integral can now be evaluated either analytically or using a numerical method (e.g., a calculator with an integration function). Upon evaluating the integral, we get: E ≈ 1542.5 N/C Hence, the electric field on the axis of the washer at a distance of 30.0 cm from the center of the washer is approximately 1542.5 N/C.

Key concepts

Electric Field Due to Ring of Charge

Understanding the electric field due to a ring of charge is crucial when studying electrostatics. Imagine a hypothetical ring with a uniform charge distribution. The electric field it generates at a point along the axis passing through the center is directed either towards or away from the ring depending on the charge's polarity. For a positively charged ring, the field points away, while for a negatively charged ring, it points towards the center.

The formula for the electric field at a distance 'x' from the center of the ring along its axis is given by:
\[ dE = \frac{k dq}{(x^2 + r^2)^{3/2}} \cdot x \]
Here, 'dq' represents a small element of charge on the ring, 'r' is the ring's radius, and 'k' is the Coulomb's constant. The x component of the electric field due to this element is proportional to the distance 'x' and inversely related to the cube of the distance from the element to the point under consideration.

To find the electric field due to the entire ring, we'd integrate over the entire circumference. In the case of the washer, which is essentially a large ring with a smaller ring removed from the center, we'll consider the contributions of each infinitesimal ring of charge to find the overall electric field.

Surface Charge Density

Surface charge density is a measure of how much charge is distributed over a given area. It's denoted by the Greek letter sigma (\( \sigma \)) and mathematically expressed as:
\[ \sigma = \frac{Q}{A} \]
where 'Q' is the total charge, and 'A' is the area over which the charge is spread. In our washer problem, the washer has a specific amount of charge spread uniformly over the top surface. To calculate the surface charge density, we use the formula above, taking into account the area of the outer circle and subtracting the area of the inner circle, corresponding to the hole. Knowing the charge density allows us to calculate the small amount of charge 'dq' on each infinitesimal ring by multiplying the surface charge density by the circumference of the ring and its infinitesimal width, giving us a better grasp on the distribution of charge for integration.

Integration in Electric Field Calculation

The process of integration is the cornerstone technique for solving complex electric field problems like the one involving our washer. Integration allows us to sum up the contributions to the electric field from each infinitesimal charge element 'dq' on the washer's surface. For a concise understanding, consider each small charge element contributing a tiny electric field 'dE'.

To find the total electric field 'E', we integrate 'dE' over the limits that correspond to the size of the washer—this means integrating from the inner radius to the outer radius.
\[ E = \int_{r_{inner}}^{r_{outer}} dE \]
Given that 'dE' involves the electric field expression due to a thin ring, the integration will include the surface charge density and will account for the inverse cube relationship of the distance between the charge element and the point of interest along the axis. Upon performing this integration, we obtain the total electric field at the specified distance from the center of the washer, completing our electric field calculation.