A solid sphere of radius R has a charge Q distributed in its volume with a charge density Δ = kr^a,

Q: Why do we use $4\pi x^2 dx$ for the volume element?

Because the charge density only depends on the radial distance $r$, we can divide the sphere into infinitesimally thin concentric spherical shells. The surface area of a shell of radius $x$ is $4\pi x^2$, and its thickness is $dx$, so its volume is $dV = 4\pi x^2 dx$.

Q: Can we just use the total charge $Q$ to find the electric field inside?

No, Gauss's Law states that the electric field at a distance $r$ inside the sphere depends *only* on the charge enclosed within that specific radius $r$, not the total charge $Q$ of the entire sphere.

Q: What if $a$ was a negative number?

If $a$ is negative, the charge density would be highest at the center and decrease outwards. The math still holds as long as $a > -3$ (otherwise the total charge would be infinite). Interestingly, if $a = -1$, the electric field inside the sphere would be constant!

Q: Why did the $4\pi k$ and $\epsilon_0$ terms disappear in the final calculation?

Because we were given a *ratio* of the electric fields at two different points. When you divide $E(R/2)$ by $E(R)$, all the constant terms cancel out, leaving only the variable $r$ terms. This is a powerful technique to simplify calculations!

Q: Is the electric field zero at the center of this sphere?

Yes! Since $E(r) \propto r^{a+1}$ and we found $a=2$, the electric field is proportional to $r^3$. At the center ($r=0$), the electric field is exactly zero, which makes sense because there is no enclosed charge at a point.

Question

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho = kr^a$, where $k$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r = \frac{R}{2}$ is $\frac{1}{8}$ times that at $r = R$, find the value of $a$.

Sigma Percentile · Accepted Answer

### The Mystery of the Non-Uniform Sphere

Imagine you are exploring a solid sphere, but this isn't your standard textbook sphere where the charge is spread out perfectly evenly. Instead, the charge density $ho$ changes as you move away from the center, following a specific mathematical rule: $ho = kr^a$.

Our mission is to find the mysterious exponent $a$. To do this, we are given a fascinating clue: the electric field exactly halfway to the surface ($r = R/2$) is exactly one-eighth of the electric field at the surface ($r = R$).

### Gauss's Law to the Rescue

When dealing with spherically symmetric charge distributions, Gauss's Law is our ultimate weapon. It tells us that the electric field $E$ at any distance $r$ from the center depends *only* on the charge enclosed within that radius.

The formula is beautifully simple:
$$E(r) = \frac{1}{4\pi\epsilon_0} \frac{q_{	ext{enclosed}}}{r^2}$$

To find the electric field, we first need to figure out exactly how much charge $q_{	ext{enclosed}}$ is trapped inside a radius $r$.

### Setting Up the Volume Integral

Because the charge density varies with $r$, we can't just multiply density by total volume. We have to build the sphere layer by layer, like an onion.

We take an infinitesimally thin spherical shell of radius $x$ and thickness $dx$. The volume of this tiny shell is its surface area multiplied by its thickness: $dV = 4\pi x^2 dx$.

The charge inside this thin shell is $dq = ho dV$. Substituting our given density $ho = kx^a$, we get:
$$dq = (kx^a)(4\pi x^2) dx$$

To find the total enclosed charge up to radius $r$, we integrate this expression from the center ($x=0$) to our radius of interest ($x=r$):
$$q(r) = \int_0^r 4\pi k x^{a+2} dx$$

### Evaluating the Enclosed Charge

Now, let's perform the integration. The constants $4\pi$ and $k$ can be pulled out of the integral.

Using the standard power rule for integration, the integral of $x^{a+2}$ becomes $\frac{x^{a+3}}{a+3}$. Evaluating this from $0$ to $r$, we get:
$$q(r) = \frac{4\pi k}{a+3} r^{a+3}$$

This tells us exactly how the enclosed charge grows as we move outward from the center!

### The Electric Field Profile

Now that we have the enclosed charge, let's plug it back into our Gauss's Law equation to find the electric field profile:
$$E(r) = \frac{1}{4\pi\epsilon_0} \frac{\frac{4\pi k}{a+3} r^{a+3}}{r^2}$$

Notice how the $r^2$ in the denominator beautifully cancels out two powers of $r$ from the numerator. We can ignore all the constants because we only care about how $E$ depends on $r$. We find a stunningly simple proportionality:
$$E(r) \propto r^{a+1}$$

### Cracking the Code

We are finally ready to use our crucial clue! We know that the electric field at $r = R/2$ is $\frac{1}{8}$ of the electric field at $r = R$.

Let's set up the ratio using our proportionality:
$$\frac{E(R/2)}{E(R)} = \frac{(R/2)^{a+1}}{R^{a+1}} = \left(\frac{1}{2}ight)^{a+1}$$

We are given that this ratio equals $\frac{1}{8}$. So, we can write:
$$\left(\frac{1}{2}ight)^{a+1} = \frac{1}{8}$$

Since we know that $\frac{1}{8}$ is simply $\left(\frac{1}{2}ight)^3$, we can equate the exponents:
$$a + 1 = 3$$

Solving this simple equation gives us our final answer:
$$a = 2$$

The charge density increases as the square of the distance from the center, meaning the charge is heavily concentrated near the outer surface of the sphere!

Enter Numerical Value:

Visualized Solution

$Visualizing the Sphere$

$Gauss’s Law$

$Enclosed Charge Setup$

$Evaluating the Integral$

$Electric Field Proportionality$

$Applying the Given Condition$

$Solving for a$

$The Way Forward$

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

The Mystery of the Non-Uniform Sphere

Gauss's Law to the Rescue

Setting Up the Volume Integral

Evaluating the Enclosed Charge

The Electric Field Profile

Cracking the Code

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Sphere

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gauss’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Enclosed Charge Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field Proportionality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Given Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for a

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward