An infinitely long solid cylinder of radius R has a uniform volume charge density \rho. It has a sph

Q: Why do we subtract the electric fields?

Because the cavity represents an absence of charge. By calculating the field of a complete cylinder, we have included the charge in the cavity region. To correct this, we subtract the field that would be produced by the charge in that cavity.

Q: How do we find the electric field of the solid cylinder?

We can use Gauss's law. For a point outside the cylinder, it behaves like a line charge with linear charge density $\lambda = \pi R^2 \rho$. The field is $E = \frac{\lambda}{2\pi\varepsilon_0 r}$.

Q: Can we use Gauss's law directly on the cylinder with the cavity?

No, because the presence of the spherical cavity breaks the cylindrical symmetry required to easily apply Gauss's law. Superposition is the best approach here.

Q: What is the charge of the spherical cavity?

The charge is the volume of the sphere multiplied by the volume charge density: $Q = \frac{4}{3}\pi (R/2)^3 \rho = \frac{\pi R^3 \rho}{6}$.

Q: Why is the distance for both fields $2R$?

The point $P$ is at a distance $2R$ from the axis of the cylinder, which is also the center of the spherical cavity. So, the distance from the center of both the cylinder and the sphere to point $P$ is $2R$.

Question

An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho$. It has a spherical cavity of radius $\frac{R}{2}$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$, which is at a distance $2R$ from the axis of the cylinder, is given by the expression $\frac{23\rho R}{16k\varepsilon_0}$. The value of $k$ is

Sigma Percentile · Accepted Answer

The principle of superposition is one of the most elegant and powerful tools in electrostatics. When faced with a complex charge distribution—like a solid object with a cavity—we can often break it down into simpler, symmetric components. In this problem, we have an infinitely long solid cylinder with a spherical cavity, and we need to find the electric field at a point $P$ outside the cylinder.

### Analyzing the Setup

We are given a solid cylinder of radius $R$ with a uniform volume charge density $ho$. Inside this cylinder, there is a spherical cavity of radius $R/2$ centered on the cylinder's axis. We need to determine the electric field at a point $P$, which lies at a distance $2R$ from the axis.

Directly applying Gauss's law to this geometry is impossible because the spherical cavity destroys the cylindrical symmetry. However, we can use the **principle of superposition**. We can treat the system as a complete, solid cylinder with charge density $ho$, superimposed with a solid sphere of charge density $-ho$ at the location of the cavity.

$$ \vec{E}_P = \vec{E}_{	ext{cylinder}} - \vec{E}_{	ext{sphere}} $$

### The Master Equation: Field of the Cylinder

First, let's calculate the electric field produced by the complete solid cylinder at point $P$. For a point outside the cylinder, we can use Gauss's law. The cylinder behaves like a line charge located at its axis.

The charge per unit length $\lambda$ of the cylinder is the volume of a unit length multiplied by the charge density:

$$ \lambda = \pi R^2 ho $$

Using the formula for the electric field of a line charge, $E = \frac{\lambda}{2\pi\varepsilon_0 r}$, where $r = 2R$:

$$ E_{	ext{cyl}} = \frac{\pi R^2 ho}{2\pi\varepsilon_0 (2R)} = \frac{ho R}{4\varepsilon_0} $$

### Field of the Spherical Cavity

Next, we calculate the electric field that would be produced by the "missing" charge in the spherical cavity. We treat this as a solid sphere of radius $R/2$ with charge density $ho$.

The total charge $Q$ of this sphere is its volume multiplied by the charge density:

$$ Q = \frac{4}{3}\pi \left(\frac{R}{2}ight)^3 ho = \frac{\pi R^3 ho}{6} $$

For a point outside the sphere, it behaves like a point charge located at its center. The electric field at distance $2R$ is:

$$ E_{	ext{sph}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{(2R)^2} = \frac{1}{4\pi\varepsilon_0} \frac{\frac{\pi R^3 ho}{6}}{4R^2} = \frac{ho R}{96\varepsilon_0} $$

### Final Calculation

Now, we apply superposition. The net electric field at $P$ is the field of the cylinder minus the field of the sphere. Both fields point radially outward along the y-axis, so we can simply subtract their magnitudes:

$$ E_P = E_{	ext{cyl}} - E_{	ext{sph}} $$

$$ E_P = \frac{ho R}{4\varepsilon_0} - \frac{ho R}{96\varepsilon_0} $$

To subtract these fractions, we find a common denominator, which is $96\varepsilon_0$:

$$ E_P = \frac{24ho R - ho R}{96\varepsilon_0} = \frac{23ho R}{96\varepsilon_0} $$

The problem states that the electric field is given by the expression $\frac{23ho R}{16k\varepsilon_0}$. Equating our result to this expression:

$$ \frac{23ho R}{96\varepsilon_0} = \frac{23ho R}{16k\varepsilon_0} $$

$$ 16k = 96 $$

$$ k = 6 $$

**The value of $k$ is 6.** This problem beautifully demonstrates how superposition can simplify seemingly intractable electrostatic configurations into manageable, atomic calculations.

Enter Numerical Value:

Visualized Solution

\text{Understanding the Geometry}

\text{Principle of Superposition}

\text{Electric Field due to Complete Cylinder}

\text{Electric Field due to Spherical Cavity}

\text{Net Electric Field at } P

\text{Simplifying the Expression}

\text{Finding the Value of } k

\text{Conclusion}

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Analyzing the Setup

The Master Equation

Field of the Spherical Cavity

Final Calculation

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