Problem 41 - A Complete Resource Book in Physics for JEE Main

Question

A cube of side \(b\) has a charge \(q\) at each of its vertices. The electric potential at the centre of the cube is (A) \(\frac{4 q}{\sqrt{3} \pi \varepsilon_{0} b}\) (B) \(\frac{\sqrt{3} q}{\pi \varepsilon_{0} b}\) (C) \(\frac{2 q}{\pi \varepsilon_{0} b}\) (D) Zero

Step-by-Step Solution

Answer

The electric potential at the center of a cube with side \(b\) and charge \(q\) at each of its vertices is \(\frac{4 q}{\sqrt{3} \pi \varepsilon_{0} b}\).

Step 1: Understanding electric potential and its formula

The electric potential at a point due to a point charge is given by the formula: \[V=\frac{kq}{r}\] Where \(k\) is the electrostatic constant, \(q\) is the charge, and \(r\) is the distance from the point charge to the point where we want to find the electric potential. In this case, the electrostatic constant \(k\) can be expressed in terms of the vacuum permittivity \(\varepsilon_0\), as \(k=\frac{1}{4\pi\varepsilon_0}\).

Step 2: Finding the distance between the center of the cube and its vertices

In order to find the electric potential at the center of the cube due to the charges at its vertices, we first need to find the distance between the center of the cube and its vertices. Let the side of the cube be \(b\). The cube consists of 8 vertices, and since the cube is symmetric, the distance from the center to each vertex will be the same. To find this distance, we can consider one vertex and the center as two diagonal points of the cube. Let's call the distance \(r\). Using the Pythagorean theorem in 3D space, we get: \[r^2 = (\frac{b}{2})^2 + (\frac{b}{2})^2 + (\frac{b}{2})^2\] Solving for \(r\), we find that: \[r = \frac{b\sqrt{3}}{2}\]

Step 3: Calculating the electric potential at the center of the cube due to all vertices

Now that we know the distance between the center of the cube and its vertices, we can find the electric potential at the center due to each vertex using the formula mentioned earlier. The total electric potential at the center will be the sum of the electric potentials due to each vertex. Since there are 8 vertices, the total electric potential will be: \[V_{total} = 8\times\frac{1}{4\pi\varepsilon_0}\times\frac{q}{\frac{b\sqrt{3}}{2}}\] Simplifying the equation, we get: \[V_{total} = \frac{8q}{4\pi\varepsilon_0\times\frac{b\sqrt{3}}{2}}\] \[V_{total} = \frac{4q}{\sqrt{3}\pi\varepsilon_0 b}\] The correct answer is: (A) \(\frac{4 q}{\sqrt{3} \pi \varepsilon_{0} b}\).