Question

A charge \(Q=\) \(+5.60 \mu C\) is uniformly distributed on a thin cylindrical plastic shell. The radius, \(R\), of the shell is \(4.50 \mathrm{~cm}\). Calculate the electric potential at the origin of the \(x y\) -coordinate system shown in the figure. Assume that the electric potential is zero at points infinitely far away from the origin.

Short answer

Answer: The electric potential at the origin is \(1.11 \times 10^{6}\,V\).

Step-by-step solution

Set up the integral

We need to convert the given values for radius and total charge in meters and coulombs respectively. Since, the radius is given in centimeters, we will convert it into meters by multiplying it by \(10^{-2}\), and we will multiply the charge with \(10^{-6}\) to convert it to coulombs: $$ R = 4.50 \times 10^{-2}\,m $$ $$ Q = 5.60 \times 10^{-6}\,C $$ A small charge element, \(dq\), is present on the surface of the cylindrical shell, and we can write the expression for potential due to this small charge element at the origin as: $$ dV = \frac{k dq}{R} $$ To obtain the total potential we integrate \(dV\) over the entire charge distribution: $$ V = \int_{0}^{Q} \frac{k}{R} dq $$

Calculate the electric potential

Now we can evaluate the integral: $$ V = \frac{k}{R} \int_{0}^{Q} dq $$ $$ V = \frac{k}{R}\left[ q\right]_0^Q $$ $$ V = \frac{k \cdot Q}{R} $$ Next, we substitute the values of \(k\), \(Q\), and \(R\) into the equation: $$ V = \frac{(8.99 \times 10^{9}\,N m^2 C^{-2})(5.60 \times 10^{-6}\,C)}{4.50 \times 10^{-2}\,m} $$

Solve for the electric potential

Now we can solve for the electric potential at the origin: $$ V = \frac{(8.99 \times 10^{9})(5.60 \times 10^{-6})}{4.50 \times 10^{-2}} = 1.11 \times 10^{6}\,V $$ Thus, the electric potential at the origin is \(1.11 \times 10^{6}\,V\).

Key concepts

Cylindrical Charge Distribution

Understanding the cylindrical charge distribution is crucial when dealing with problems in electrostatics, especially when the symmetry of the problem suggests a cylindrical geometry. This type of charge distribution is characterized by a uniform charge over the surface of a cylinder.

For example, in the exercise provided, the charge is uniformly distributed on a thin cylindrical shell. This symmetry allows us to use the concept of a charge element, denoted as \(dq\), which represents a small segment of the total charge \(Q\). It's important to understand that \(dq\) is not just a smaller charge, but an infinitesimally small portion of \(Q\), enabling us to integrate over the entire surface to find the total effect.

In our case, we need to calculate the electric potential at the origin due to this distribution. Since the charge is uniformly distributed, every charge element contributes equally to the potential at the origin. The symmetry of the cylinder allows us to simplify the problem, as we don't need to worry about the angle around the cylinder—only the distance to the origin and the total charge matter. Recognizing these symmetries is fundamental to solving these problems efficiently and accurately.

Integral Calculus in Physics

Integral calculus is a powerful mathematical tool often used in physics to solve problems involving quantities that are distributed over space or time. The technique of integrating lets us sum up infinitely small contributions to obtain a total quantity.

In the context of the electric potential problem, we applied integral calculus to sum up the small contributions of potential \(dV\) from each infinitesimal charge element \(dq\) along the cylindrical shell. The integral sign \(\int\) denotes this 'summing' process, and the limits of integration (from 0 to \(Q\)) specify over what range we are summing.

The integration process simplifies due to the constant distance \(R\) and the constant \(k\) (Coulomb's constant), resulting in the integral of \(dq\) which is simply \(Q\) after evaluating the definite integral. It's essential for students to grasp how integrals allow us to move from a differential element to the total quantity. Integral calculus is a cornerstone in physics problems when continuous distributions, like charge, mass, or probability, are involved.

Coulomb's Law

Coulomb's law is a fundamental principle in electrostatics that describes the force between two point charges. The law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this exercise, we apply the principle of Coulomb's Law to calculate electric potential, which describes the work done to bring a unit positive charge from infinity to a point within an electric field. The formula derived from Coulomb's Law for a point charge \(q\) is given as \(V = k \cdot q / r\), where \(V\) is the electric potential, \(k\) is the Coulomb constant approximately equal to \(8.99 \times 10^{9} N m^2 C^{-2}\), \(q\) is the charge, and \(r\) is the distance from the charge to the point of interest.

In the context of a uniformly charged cylinder, due to symmetry, we consider the entire surface at a constant distance from the origin, and the law simplifies the relationship for this configuration. We integrate the small potentials \(dV\) due to each infinitesimal charge to get the total potential at the origin. Coulomb's Law is integral to understanding how charges interact and is foundational for solving problems that involve electrostatic forces and fields.