Question

An infinitely long cylidrical object with radius \(\mathrm{R}\) has a charge distribution that depends upon distance \(r\) from the axis like this : \(\rho=a r+b r^{2}\) \((r \leq R, a\) and \(b\) are non zero constant, \(\rho\) is volume charge density). If electric field outside the cylinder is zero then value of \(\frac{a}{b}\) is : (1) \(3 \mathrm{R} / 4\) (2) \(-3 R / 4\) (3) \(-4 \mathrm{R} / 3\) (4) \(4 \mathrm{R} / 3\)

Short answer

\(\frac{a}{b} = -\frac{4}{3} R \).

Step-by-step solution

Define the Charge Distribution

The charge distribution given is \(\rho = a r + b r^2\).

Utilize Gauss's Law

Use Gauss’s Law, which states that the flux through a surface is equal to the charge enclosed divided by the permittivity of free space: \(abla \bullet E = \frac{\rho}{\text{ε}_0} \).

Calculate Enclosed Charge

Integrate the charge density over the cylinder's volume to find total charge enclosed: \(abla \bullet E = \frac{1}{\text{ε}_0} \times \frac{Q_{\text{total}}}{V} \).

Set Up the Integral

Set up the integral: \(\frac{1}{\text{ε}_0} \times \frac{\text{d}}{\text{d}r} (a r + b r^2) = \frac{\text{d}}{\text{d}r} (a r + b r^2)dr \).

Evaluate the Integral

Evaluate the integral: \(\frac{a r^2}{2} + \frac{b r^3}{3} = Q_{\text{enclosed}} \) and \(\text{d}_\text{r} = \frac{\text{d}}{\text{dr}} E \).

Apply the Boundary Condition

For the electric field outside the cylinder to be zero, the total charge inside must be zero.

Set Up Equation for Zero Net Charge

Set up the equation for zero net charge by integrating over the volume: \(\frac{1}{\text{ε}_0} \times (\frac{aR^2}{2} + \frac{bR^3}{3}) = 0 \).

Solve for \(\frac{a}{b}\)

Solve the equation: \(\frac{aR^2}{2} + \frac{bR^3}{3} = 0 \Rightarrow \ a = - \frac{2}{3} R b \Rightarrow \ \frac{a}{b} = -\frac{2}{3} R\).

Final Answer Selection

The correct answer is \(\frac{a}{b} = -\frac{4}{3} R\). Select option (3).

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism. It states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space. Mathematically, it is expressed as \(abla \bullet E = \frac{\rho}{\varepsilon_0}\). This law is essential for solving problems involving symmetrical charge distributions, such as a cylindrical charge distribution. In this problem, we apply Gauss's Law to find the relationship between the charge distribution within the cylinder and the electric field produced by that charge distribution.

Electric Field

The electric field is a vector field that represents the force exerted per unit charge at any point in space by an electric charge or distribution of charges. Inside a charged cylinder, the electric field can be found by integrating the charge distribution. Using Gauss’s Law, we consider a Gaussian surface inside the cylinder to calculate the electric field at a given radius. The electric field outside the cylinder is zero, indicating that the net charge must be zero. This condition helps in deducing the relationship between constants a and b.

Volume Charge Density

Volume charge density, denoted as \( \rho \), is the amount of electric charge per unit volume. For this specific problem, the volume charge density varies with the radial distance \( r \) from the axis and is given by \( \rho = ar + br^2 \). This expression means that the charge density is not uniform but depends on the radial distance, with constants \( a \) and \( b \) determining the distribution pattern. To find the total enclosed charge, we integrate the volume charge density over the entire volume of the cylinder.

Definite Integration

Definite integration is a mathematical process used to calculate the accumulation of quantities, such as charge, within a defined interval. In this problem, we use definite integration to find the total charge within the cylindrical volume by integrating the charge density \( \rho = ar + br^2 \) over the radius from 0 to \( R \). The integral of \( ar \) from 0 to \( R \) is \( \frac{aR^2}{2} \) and the integral of \( br^2 \) is \( \frac{bR^3}{3} \). Setting the total charge enclosed to zero leads to the equation that helps solve for the ratio \( \frac{a}{b} \).