Question

A point charge Q is placed at extension of diagonal of a cube of side a, very close to the vertex. The magnitude of sum of electric flux through the sides of the cube which met at this vertex is

Answer 1

\frac{Q}{8\epsilon_0}

Answer 2

The problem asks for the sum of electric flux through the three faces of a cube that meet at a specific vertex, given that a point charge Q is placed very close to this vertex, on the extension of the cube's main diagonal.

Let's denote the vertex closest to the charge as V. Let the cube be placed in a coordinate system such that V is at the origin (0,0,0). The cube then occupies the region $0 \le x \le a$, $0 \le y \le a$, $0 \le z \le a$. The main diagonal of the cube extends from (0,0,0) to (a,a,a). The problem states that the charge Q is placed on the extension of this diagonal, very close to the vertex (0,0,0). This means the charge Q is at a point P, say $(-k, -k, -k)$, where k is a very small positive number ($k \to 0^+$).

The three faces of the cube that meet at the vertex V(0,0,0) are:

1. Face 1: $x=0$, with normal vector $d\vec{A}_1 = -dy dz \hat{i}$ (outward normal).
2. Face 2: $y=0$, with normal vector $d\vec{A}_2 = -dx dz \hat{j}$ (outward normal).
3. Face 3: $z=0$, with normal vector $d\vec{A}_3 = -dx dy \hat{k}$ (outward normal).

The other three faces are: 4. Face 4: $x=a$, with normal vector $d\vec{A}_4 = dy dz \hat{i}$. 5. Face 5: $y=a$, with normal vector $d\vec{A}_5 = dx dz \hat{j}$. 6. Face 6: $z=a$, with normal vector $d\vec{A}_6 = dx dy \hat{k}$.

According to Gauss's Law, the total electric flux through a closed surface is $\Phi_{total} = Q_{enc} / \epsilon_0$. Since the charge Q is outside the cube, the total flux through the cube is zero: $\Phi_{total} = \Phi_1 + \Phi_2 + \Phi_3 + \Phi_4 + \Phi_5 + \Phi_6 = 0$.

Due to the symmetry of the problem (charge Q is on the extension of the main diagonal, and the cube is symmetric with respect to this diagonal), the electric flux through the three faces meeting at the vertex V must be equal. Let this flux be $\Phi_V$. So, $\Phi_1 = \Phi_2 = \Phi_3 = \Phi_V$. Similarly, the flux through the three opposite faces (Face 4, Face 5, Face 6) must also be equal. Let this flux be $\Phi_{Opp}$. So, $\Phi_4 = \Phi_5 = \Phi_6 = \Phi_{Opp}$.

Substituting these into the total flux equation: $3\Phi_V + 3\Phi_{Opp} = 0$ This implies $\Phi_{Opp} = -\Phi_V$.

The question asks for the sum of electric flux through the sides of the cube which met at this vertex, which is $3\Phi_V$.

Consider an imaginary scenario where we construct a larger cube of side length $2a$ centered at the origin (0,0,0), with the charge Q at its center. This larger cube would consist of 8 smaller cubes, each of side length $a$. If the charge Q were placed at the vertex (0,0,0), it would be shared by 8 such identical cubes. The total flux through one such cube would be $Q/(8\epsilon_0)$. In this case, the charge Q is at $(-k,-k,-k)$, which is infinitesimally outside the cube $0 \le x,y,z \le a$.

Let's use the solid angle concept. The flux through a surface due to a point charge Q is given by $\Phi = \frac{Q}{4\pi\epsilon_0} \Omega$, where $\Omega$ is the solid angle subtended by the surface at the charge. The solid angle subtended by one octant of space at its origin is $\frac{1}{8}$ of the total solid angle ($4\pi$), which is $\frac{4\pi}{8} = \frac{\pi}{2}$. If the charge Q were at the vertex V(0,0,0), the solid angle subtended by the cube at Q (from inside the cube) is $\pi/2$. The flux through the cube would be $Q/(8\epsilon_0)$. The faces $x=0, y=0, z=0$ (the ones meeting at V) have the charge on their boundary. For a charge on the boundary, the flux through that specific boundary surface is often considered zero in such problems, or more rigorously, the field is singular. However, if the charge is infinitesimally displaced, the flux is non-zero.

When the charge Q is at $(-k,-k,-k)$, it is outside the cube. The flux through the three faces meeting at V ($x=0, y=0, z=0$) will be negative (field lines entering the cube). The flux through the three opposite faces ($x=a, y=a, z=a$) will be positive (field lines exiting the cube).

Consider the solid angle subtended by the three faces meeting at the vertex V (i.e., $x=0, y=0, z=0$) at the charge Q located at $(-k,-k,-k)$. As $k \to 0$, the charge approaches the vertex V. The three faces meeting at V effectively cover a solid angle of $3 \times \frac{1}{2} \times \frac{\pi}{2} = \frac{3\pi}{4}$? No, this is not correct.

Let's consider the scenario from the perspective of the charge Q. The charge Q is at $P(-k,-k,-k)$. The vertex V is at $(0,0,0)$. The three faces $x=0, y=0, z=0$ are "in front" of the charge. The solid angle subtended by the entire infinite plane $x=0$ at P is $2\pi$. The solid angle subtended by a quarter plane (e.g., $x=0, y \ge 0, z \ge 0$) at P is $\pi/2$. The three faces $x=0, y=0, z=0$ form a "corner" of the cube. The solid angle subtended by the entire cube at the charge Q (since Q is outside) is 0.

Let's consider the problem in reverse. If the charge Q were inside the cube, at the vertex (0,0,0). Then the flux through the cube would be $Q/(8\epsilon_0)$. The flux through the three faces meeting at the vertex (0,0,0) (i.e., $x=0, y=0, z=0$) would be zero, as the electric field lines are tangential to these surfaces at the vertex, or more accurately, the charge lies on these surfaces. Then, all the flux $Q/(8\epsilon_0)$ must pass through the other three faces ($x=a, y=a, z=a$). By symmetry, each of these three faces would have a flux of $(Q/(8\epsilon_0))/3 = Q/(24\epsilon_0)$.

Now, consider the charge Q at $(-k,-k,-k)$ and the cube at $0 \le x,y,z \le a$. The total flux is zero. The three faces meeting at V(0,0,0) are $F_1(x=0), F_2(y=0), F_3(z=0)$. The three opposite faces are $F_4(x=a), F_5(y=a), F_6(z=a)$. By symmetry, $\Phi_1 = \Phi_2 = \Phi_3$ and $\Phi_4 = \Phi_5 = \Phi_6$. Also, $\Phi_1 + \Phi_4 = 0$, $\Phi_2 + \Phi_5 = 0$, $\Phi_3 + \Phi_6 = 0$. This is not necessarily true for individual pairs of faces. However, it is true that $\Phi_1 = -\Phi_4$ (as if the charge was at the center of a large cube formed by two cubes, one with $x<0$ and one with $x>0$).

Let's consider the solid angle approach more carefully. The total solid angle around the charge Q is $4\pi$. The solid angle subtended by the "inner" part of the vertex (the part that would be inside the cube if Q were at the vertex) is $\pi/2$. The solid angle subtended by the "outer" part of the vertex (the part where Q is located) is $4\pi - \pi/2 = 7\pi/2$. The charge Q is in the region $x<0, y<0, z<0$. The three faces of the cube meeting at (0,0,0) are $x=0, y=0, z=0$. These faces are "visible" from Q. The solid angle subtended by the cube at the charge Q is zero (since Q is outside).

This is a classic problem that often relies on a conceptual understanding of solid angles and symmetry. When the charge is placed very close to a vertex, on the extension of the diagonal, it implies that the field lines effectively originate from a point just outside the cube. The electric field lines will enter the cube through the three faces meeting at that vertex. The electric field lines will exit the cube through the three opposite faces. Since the charge is placed symmetrically with respect to the diagonal, the flux through the three faces meeting at the vertex will be equal in magnitude (and negative, as flux enters). Similarly, the flux through the three opposite faces will be equal in magnitude (and positive, as flux exits).

Let $\Phi_{corner}$ be the sum of flux through the three faces meeting at the vertex V. Let $\Phi_{opposite}$ be the sum of flux through the three faces opposite to V. Since the charge is outside, $\Phi_{corner} + \Phi_{opposite} = 0$. So, $\Phi_{corner} = -\Phi_{opposite}$.

Consider the total solid angle $4\pi$ around the charge. If the charge is at the vertex, the cube occupies $1/8$ of the solid angle ($4\pi/8 = \pi/2$). The flux is $Q/(8\epsilon_0)$. The remaining $7/8$ of the solid angle ($7\pi/2$) is outside the cube. When the charge is at $(-k,-k,-k)$, the solid angle subtended by the entire cube at the charge is 0. However, the solid angle subtended by the three faces meeting at the vertex V(0,0,0) at the charge Q($-k,-k,-k$) is $\frac{1}{8}$ of the total solid angle ($4\pi$) if we consider the charge to be at the center of a larger cube formed by 8 smaller ones. The problem is tricky because the charge is outside.

Let's consider the problem from a different perspective. Imagine the charge Q is at the origin (0,0,0). Then the flux through an infinite plane (e.g., $x=a$) is $Q/(2\epsilon_0)$ if the charge is on one side. However, this is not a plane, but a finite face.

The key phrase is "very close to the vertex". This means that the distance from Q to any point on the three faces meeting at the vertex is much smaller than the side length 'a', except for points far from the vertex. For points very close to the vertex, the field lines are almost radial from the vertex. The solid angle subtended by the three faces meeting at V, at the charge Q, is exactly the solid angle subtended by a corner of a cube. If the charge Q is at the vertex (0,0,0), then the flux through the faces meeting at (0,0,0) is zero. The flux through the other three faces is $Q/(8\epsilon_0)$. If the charge Q is at $(-k,-k,-k)$, then it's infinitesimally shifted. The flux through the three faces meeting at (0,0,0) will not be zero. Consider the case where Q is inside a very small sphere around the vertex. The total flux through that small sphere is $Q/\epsilon_0$. The fraction of this flux that passes through the faces of the cube depends on the solid angle. The solid angle subtended by the three faces meeting at the vertex (0,0,0) from the point Q(-k,-k,-k) is $\frac{1}{8}$ of the total solid angle ($4\pi$) for the "outward" direction from the charge, i.e., towards the cube. So, the solid angle is $\pi/2$. The flux entering the cube through these three faces is $\Phi_{V,faces} = -\frac{Q}{4\pi\epsilon_0} \frac{\pi}{2} = -\frac{Q}{8\epsilon_0}$. The negative sign indicates the flux is entering the cube. The magnitude of the sum of electric flux through these sides is $|-Q/(8\epsilon_0)| = Q/(8\epsilon_0)$.

Final check: If Q is at $(-k,-k,-k)$, it is outside the cube. So total flux through the cube must be zero. The flux through the three faces meeting at (0,0,0) (the ones facing the charge) is $\Phi_{in}$. The flux through the three faces opposite to (0,0,0) (the ones away from the charge) is $\Phi_{out}$. $\Phi_{in} + \Phi_{out} = 0$. By symmetry, the solid angle subtended by the three faces meeting at (0,0,0) at the charge Q(-k,-k,-k) is $\pi/2$. So, $\Phi_{in} = -\frac{Q}{4\pi\epsilon_0} \frac{\pi}{2} = -\frac{Q}{8\epsilon_0}$. The question asks for the magnitude of the sum of electric flux through the sides of the cube which met at this vertex. This is $|\Phi_{in}|$.

Magnitude of sum of electric flux = $\left|-\frac{Q}{8\epsilon_0}\right| = \frac{Q}{8\epsilon_0}$.

The final answer is $\boxed{\frac{Q}{8\epsilon_0}}$.

Explanation:

1. The charge Q is placed very close to a vertex of the cube, on the extension of its main diagonal. This means the charge is outside the cube.
2. According to Gauss's Law, the total electric flux through a closed surface (like the cube) with no charge enclosed is zero.
3. Let the vertex be V. The three faces meeting at V are $F_1, F_2, F_3$. The other three faces are $F_4, F_5, F_6$.
4. Due to the symmetry of the charge placement on the diagonal extension, the flux through $F_1, F_2, F_3$ will be equal. Let this be $\Phi_V$. Similarly, the flux through $F_4, F_5, F_6$ will be equal. Let this be $\Phi_{Opp}$.
5. The total flux is $3\Phi_V + 3\Phi_{Opp} = 0$, which implies $\Phi_{Opp} = -\Phi_V$.
6. The flux through a surface due to a point charge is given by $\Phi = \frac{Q}{4\pi\epsilon_0} \Omega$, where $\Omega$ is the solid angle subtended by the surface at the charge.
7. When a charge is placed very close to a vertex, the three faces meeting at that vertex effectively subtend a solid angle of $\frac{1}{8}$ of the total solid angle ($4\pi$) from the perspective of the charge, if we consider the charge to be at the center of an imaginary larger cube formed by 8 identical smaller cubes. This solid angle is $\frac{1}{8} \times 4\pi = \frac{\pi}{2}$.
8. Since the charge is outside the cube and on the extension of the diagonal leading to the vertex, the electric field lines will be entering the cube through these three faces. Hence, the flux will be negative.
9. The sum of electric flux through these three faces is $\Phi_{V,sum} = -\frac{Q}{4\pi\epsilon_0} \left(\frac{\pi}{2}\right) = -\frac{Q}{8\epsilon_0}$.
10. The magnitude of this sum is $\left|-\frac{Q}{8\epsilon_0}\right| = \frac{Q}{8\epsilon_0}$.

The final answer is $\boxed{\frac{Q}{8\epsilon_0}}$.

The final answer is $\boxed{\frac{Q}{8\epsilon_0}}$