Animated Solution for Physics - Electrostatics: A point charge $+Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct?

A Charge on the Edge Imagine a beautifully symmetric hemispherical surface of radius $R$. Now, place a point charge $+Q$ just outside its top pole. This seemingly simple setup is a classic playground for testing your intuition about Gauss's Law, electric flux, and potential. Let's break down the physics statement by statement to uncover the hidden elegance of this problem.

Option (c). What is the total electric flux passing through this entire closed surface (the curved dome plus the flat base)?

Gauss's Law states that the net electric flux through any closed surface is strictly proportional to the charge enclosed within it:

Since our charge $+Q$ is placed just outside the hemisphere, the total enclosed charge $Q_{\text{in}}$ is exactly zero. Therefore, the net flux through the entire closed surface must be zero, not $\frac{Q}{{\epsilon }_0}$. This immediately tells us that Option (c) is incorrect.

How far is any point on this circular edge from our charge $+Q$?

The charge is at a vertical distance $R$ from the center of the flat base, and the radius of the base itself is $R$. Using the Pythagorean theorem, the distance $d$ from the charge to any point on the circumference is:

Because this distance is perfectly constant for all points along the edge, the electric potential $V$ at the circumference is also constant:

Since the potential doesn't change, the circumference is indeed an equipotential line. Option (d) is absolutely correct!

What about the electric field on the flat surface itself (Option b)? Is its normal component constant?

As you move from the center of the flat base towards its edge, two things happen: 1. The distance $r$ from the charge increases (from $R$ to $\sqrt{2}R$), which means the magnitude of the electric field $E$ weakens. 2. The angle $\alpha$ that the electric field lines make with the normal to the surface changes.

The normal component is given by $E_{\perp }=E\cos \alpha$. Because both the field strength and the angle vary across the surface, their product cannot magically remain constant. Thus, Option (b) is incorrect.

Finally, let's tackle the most mathematically beautiful part of the problem: the flux through the curved surface (Option a).

From our Gauss's Law analysis, we know the total flux is zero. This implies a perfect balance:

To find ${\Phi }_{\text{flat}}$, we use the concept of a solid angle. The flat base forms a cone with its apex at the charge $+Q$. The semi-vertical angle $\theta$ of this cone is:

The solid angle $\Omega$ subtended by this cone is:

The flux through the flat surface is simply the total flux emitted by the charge ($\frac{Q}{{\epsilon }_0}$) multiplied by the fraction of the total spherical solid angle ($4\pi$) that the cone occupies:

Since ${\Phi }_{\text{curved}}=-{\Phi }_{\text{flat}}$, we get:

The negative sign is not a typo! It perfectly captures the physical reality that the electric field lines from the charge $+Q$ are entering the curved surface (inward flux is negative) and exiting the flat surface (outward flux is positive). Option (a) is brilliantly correct.

Final Answer: The correct statements are (a) and (d).