Question

An open hemisphere is centred at origin. Point A and B are on the y-axis at equal distances from the origin. When a point charge q is kept at B, flux from the hemi sphere is $\phi$. If the point charge is shifted at A then flux from the hemi sphere will be :-

• A. $\phi$
• B. $\frac{q}{\epsilon_0}-\phi$
• C. $\frac{q}{\epsilon_0}-2\phi$
• D. Zero

Answer 1

Answer: (B)

(2) $\frac{q}{\epsilon_0}-\phi$

Answer 2

To solve this problem, we will use Gauss's Law and the principle of symmetry.

Let the open hemisphere be denoted by $S_{curved}$ and its flat circular base be denoted by $S_{base}$. Together, $S_{curved}$ and $S_{base}$ form a closed spherical surface $S_{closed}$. The origin is the center of this hemisphere. Points A and B are on the y-axis at equal distances from the origin, meaning they are symmetric with respect to the plane of the base ($xy$-plane).

Case 1: Point charge $q$ is kept at B.

Let the coordinates of B be $(0, -y_0)$, and the origin be $(0,0)$. The hemisphere is centered at the origin, and from the diagram, it's the upper hemisphere (i.e., its curved surface is in the region $y > 0$). Therefore, point B is outside the closed surface formed by the hemisphere and its base.

According to Gauss's Law, the total electric flux through a closed surface enclosing no charge is zero.
So, $\Phi_{closed} = \Phi_{S_{curved}} + \Phi_{S_{base}} = 0$.

We are given that the flux from the hemisphere (meaning the curved surface) is $\phi$.
So, $\Phi_{S_{curved}} = \phi$.

Substituting this into the Gauss's Law equation:
$\phi + \Phi_{S_{base}} = 0$
$\Phi_{S_{base}} = -\phi$

Here, $\Phi_{S_{base}}$ is the flux through the flat circular base due to the charge $q$ at B. For a closed surface, the area vector for the base points outwards, which is downwards (in the negative y-direction). Since the charge $q$ at B is below the base, the electric field lines from B generally point upwards. Thus, the dot product $\vec{E} \cdot d\vec{S}_{base}$ will be negative, meaning $\Phi_{S_{base}}$ is negative. This is consistent with $\Phi_{S_{base}} = -\phi$, implying $\phi$ is a positive value.

Case 2: Point charge $q$ is shifted to A.

Let the coordinates of A be $(0, y_0)$. From the diagram, point A is inside the closed surface formed by the hemisphere and its base.

According to Gauss's Law, the total electric flux through a closed surface enclosing a charge $q$ is $\frac{q}{\epsilon_0}$.
Let the new flux through the curved surface be $\phi'$.
Let the new flux through the base be $\Phi'_{S_{base}}$.

So, $\Phi'_{closed} = \phi' + \Phi'_{S_{base}} = \frac{q}{\epsilon_0}$.

Now, we need to determine $\Phi'_{S_{base}}$. This is the flux through the flat circular base due to the charge $q$ at A.
Since A is above the base and B is below the base, and they are equidistant from the origin (symmetric with respect to the base plane), the magnitude of the flux through the base will be the same whether the charge is at A or at B.
$|\Phi'_{S_{base}}| = |\Phi_{S_{base}}|$.

From Case 1, we found $\Phi_{S_{base}} = -\phi$, so $|\Phi_{S_{base}}| = \phi$.
Therefore, $|\Phi'_{S_{base}}| = \phi$.

Now, let's consider the sign of $\Phi'_{S_{base}}$. The charge $q$ at A is above the base, so the electric field lines from A generally point downwards. The area vector for the base (for the closed surface) also points downwards. Since $\vec{E}$ and $d\vec{S}_{base}$ are generally in the same direction, their dot product $\vec{E} \cdot d\vec{S}_{base}$ will be positive.
Thus, $\Phi'_{S_{base}}$ is positive.

Combining the magnitude and sign, we get $\Phi'_{S_{base}} = \phi$.

Substitute this value into the Gauss's Law equation for Case 2:
$\phi' + \phi = \frac{q}{\epsilon_0}$
$\phi' = \frac{q}{\epsilon_0} - \phi$

The flux from the hemisphere (curved surface) when the charge is at A will be $\frac{q}{\epsilon_0} - \phi$.

The final answer is $\boxed{\text{(2)}}$.

Explanation of the solution:

1. Form a closed surface: Imagine closing the open hemisphere with a flat circular base. This forms a closed Gaussian surface.
2. Apply Gauss's Law (Initial case): When charge $q$ is at B (outside the closed surface), the total flux through the closed surface is zero.
   • Flux through curved surface ($\phi$) + Flux through base ($\Phi_{base}$) = 0.
   • This gives $\Phi_{base} = -\phi$.
3. Apply Gauss's Law (Final case): When charge $q$ is at A (inside the closed surface), the total flux through the closed surface is $q/\epsilon_0$.
   • New flux through curved surface ($\phi'$) + New flux through base ($\Phi'_{base}$) = $q/\epsilon_0$.
4. Symmetry argument for base flux: Points A and B are equidistant from the origin and symmetric with respect to the base plane. The magnitude of flux through the base is the same for charges at A or B. However, the direction of field lines relative to the outward normal of the base is opposite. From B, field lines go up (normal down, so negative flux). From A, field lines go down (normal down, so positive flux). Therefore, $\Phi'_{base} = |\Phi_{base}| = \phi$.
5. Calculate new curved surface flux: Substitute $\Phi'_{base} = \phi$ into the Gauss's Law equation for the final case:
   • $\phi' + \phi = q/\epsilon_0$
   • $\phi' = q/\epsilon_0 - \phi$.