Question

In space, a region centered at origin contains charge such that electric potential (V) varies with radial distance (r) as V = ar³ + ar². The magnitude of charge density (ρ) at radial distance r = 1 unit is Nα∈₀ then N is ______. (∈₀ is electric permittivity)

Answer 1

18

Answer 2

The electric potential is given by $V = ar^3 + ar^2$. The electric field $\vec{E}$ is related to the potential by $\vec{E} = -\nabla V$. In spherical coordinates, for a potential that depends only on the radial distance r, the electric field is radial and its magnitude is given by $E_r = -\frac{dV}{dr}$. $E_r = -\frac{d}{dr}(ar^3 + ar^2) = -(3ar^2 + 2ar)$.

The charge density $\rho$ is related to the electric field by Gauss's law in differential form, $\nabla \cdot \vec{E} = \frac{\rho}{\in_0}$. In spherical coordinates, for a spherically symmetric electric field $\vec{E} = E_r(r) \hat{r}$, the divergence is $\nabla \cdot \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 E_r)$. First, calculate $r^2 E_r$: $r^2 E_r = r^2 (-3ar^2 - 2ar) = -3ar^4 - 2ar^3$. Next, calculate $\frac{\partial}{\partial r}(r^2 E_r)$: $\frac{\partial}{\partial r}(-3ar^4 - 2ar^3) = -12ar^3 - 6ar^2$. Now, calculate the divergence: $\nabla \cdot \vec{E} = \frac{1}{r^2}(-12ar^3 - 6ar^2) = -12ar - 6a$.

Using Gauss's law, the charge density is: $\rho(r) = \in_0 (\nabla \cdot \vec{E}) = \in_0 (-12ar - 6a)$.

We need to find the magnitude of the charge density at radial distance r = 1 unit. $\rho(r=1) = \in_0 (-12a(1) - 6a) = \in_0 (-12a - 6a) = -18a\in_0$. The magnitude of the charge density at r=1 is $|\rho(r=1)| = |-18a\in_0| = 18|a|\in_0$.

The problem states that the magnitude of charge density at radial distance r = 1 unit is Nα∈₀. So, we have the equation: $18|a|\in_0 = Nα∈₀$.

For N to be a specific numerical value independent of the constants 'a' and 'α', it is implied that the constant 'a' used in the potential function is the same as the constant 'α' mentioned in the expression for the magnitude of charge density. Let's assume $a = \alpha$. Then the equation becomes: $18|\alpha|\in_0 = Nα∈₀$.

The expression Nα∈₀ represents the magnitude of the charge density, so it must be non-negative. Since $\in_0$ is positive, Nα must be non-negative. If $\alpha > 0$, then $|\alpha| = \alpha$. The equation is $18\alpha\in_0 = N\alpha\in_0$. Assuming $\alpha \neq 0$ and $\in_0 \neq 0$, we can divide by $\alpha\in_0$ to get $18 = N$. If $\alpha < 0$, then $|\alpha| = -\alpha$. The equation is $18(-\alpha)\in_0 = N\alpha\in_0$. Dividing by $\alpha\in_0$ (assuming $\alpha \neq 0$ and $\in_0 \neq 0$), we get $-18 = N$. However, if $\alpha < 0$, then Nα must be non-negative, which requires N to be non-positive. $N = -18$ fits this condition. If $\alpha = 0$, then $V=0$, $\rho=0$, and the magnitude is 0. $N\alpha\in_0 = 0$, which is true for any N. But the problem describes a region containing charge, implying $\rho \neq 0$, so $a \neq 0$, and thus $\alpha \neq 0$.

Given that the question asks for a single integer value of N without specifying the sign of α, the most likely interpretation is that N is the positive factor multiplying $|\alpha|\in_0$, or that α is assumed to be positive, or that the problem implicitly defines α as $|a|$. If the magnitude is $N|\alpha|\in_0$, then $18|a|\in_0 = N|\alpha|\in_0$. If $a=\alpha$, then $18|\alpha|\in_0 = N|\alpha|\in_0$, which gives $N=18$ (provided $\alpha \neq 0$). This yields a unique positive integer value for N.

Let's assume the convention that the constant 'a' in the potential is the same as 'α' in the expression for magnitude, and the magnitude is given as $N|\alpha|\in_0$. Then $18|\alpha|\in_0 = N|\alpha|\in_0$. $N = 18$.

Alternatively, if the problem statement "magnitude of charge density (ρ) at radial distance r = 1 unit is Nα∈₀" is interpreted strictly, it means $|\rho(1)| = N\alpha\in_0$. We found $|\rho(1)| = 18|a|\in_0$. So, $18|a|\in_0 = N\alpha\in_0$. If we assume $a=\alpha$ and $\alpha > 0$, then $18\alpha\in_0 = N\alpha\in_0$, so $N=18$. If we assume $a=\alpha$ and $\alpha < 0$, then $18(-\alpha)\in_0 = N\alpha\in_0$, so $-18 = N$. In this case, $N\alpha = (-18)(-\beta) = 18\beta = 18|\alpha|$, which is positive, consistent with the magnitude. However, integer type questions usually expect a non-negative integer unless otherwise specified. The most standard interpretation in such physics problems is that the constant α is positive or that N represents the positive numerical factor. Let's assume N is the positive numerical factor such that the magnitude is $N|a|\in_0$ where $a$ is the constant in the potential, and the problem uses α to denote this constant $a$. So, $18|a|\in_0 = N|a|\in_0$. This gives $N=18$.

The final answer is $\boxed{18}$.

Explanation of the solution:

1. Calculate the electric field $E_r(r)$ from the given potential $V(r) = ar^3 + ar^2$ using $E_r = -\frac{dV}{dr}$.
2. Calculate the charge density $\rho(r)$ from the electric field using Gauss's law in differential form $\rho = \in_0 \nabla \cdot \vec{E}$. For a spherically symmetric field, $\nabla \cdot \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 E_r)$.
3. Evaluate the charge density at r=1 unit: $\rho(1)$.
4. Calculate the magnitude of the charge density at r=1 unit: $|\rho(1)|$.
5. Equate this magnitude to the given expression $Nα∈₀$ and solve for N, assuming the constant 'a' in the potential is the same as 'α' in the expression for the magnitude, and N is the positive numerical factor.

Calculation: $V = ar^3 + ar^2$ $E_r = -\frac{d}{dr}(ar^3 + ar^2) = -(3ar^2 + 2ar)$ $\rho(r) = \in_0 \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 E_r) = \in_0 \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 (-3ar^2 - 2ar))$ $\rho(r) = \in_0 \frac{1}{r^2} \frac{\partial}{\partial r}(-3ar^4 - 2ar^3) = \in_0 \frac{1}{r^2}(-12ar^3 - 6ar^2) = \in_0 (-12ar - 6a)$ At r=1, $\rho(1) = \in_0 (-12a - 6a) = -18a\in_0$. Magnitude $|\rho(1)| = |-18a\in_0| = 18|a|\in_0$. Given magnitude is $Nα∈₀$. Assuming $a=\alpha$, magnitude is $18|\alpha|\in_0$. Equating $18|\alpha|\in_0 = Nα∈₀$. If we interpret $Nα∈₀$ as $N|\alpha|\in_0$, then $18|\alpha|\in_0 = N|\alpha|\in_0$, which gives $N=18$. If we interpret $Nα∈₀$ literally, then $18|\alpha|\in_0 = Nα∈₀$. If $\alpha>0$, $18\alpha\in_0 = N\alpha\in_0 \implies N=18$. If $\alpha<0$, $18(-\alpha)\in_0 = N\alpha\in_0 \implies -18\alpha\in_0 = N\alpha\in_0 \implies N=-18$. Since the question asks for N as a positive integer often the case in "N is ____" questions, and magnitude is positive, the standard interpretation leads to N=18.