Problem 44 - A Complete Resource Book in Physics for JEE Main

Question

Seven point charges each of charge \(q\) is placed at the seven corners of a cube of side \(a\) (one corner is empty). Find the magnitude of electric field at centre of cube. (A) Zero (B) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{a^{2}}\) (C) \(\frac{1}{3 \pi \varepsilon_{0}} \frac{q}{a^{2}}\) (D) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{7 q}{a^{2}}\)

Step-by-Step Solution

Answer

The magnitude of the electric field at the center of the cube is \(\boxed{E_{total}= \frac{1}{3\pi\varepsilon_0} \frac{q}{a^2}}\), which corresponds to option (C).

Step 1: Understand the problem

We have a cube with side length a, and seven point charges, each of magnitude q, placed at its corners. Our goal is to find the magnitude of the electric field at the center of the cube.

Step 2: Calculate electric field due to a single charge

Coulomb's Law states that the electric force between two point charges is given by \[F = k \frac{q_1q_2}{r^2}\] Here, q1 and q2 are the magnitudes of the charges, k is Coulomb's constant (\(k = \frac{1}{4\pi\varepsilon_0}\)), and r is the distance between the charges. The electric field due to a single charge can be calculated as \[E = \frac{F}{q} = \frac{kq}{r^2}\] Each charge is at the same distance from the center of the cube, which is equal to the distance across one of the cube's smaller triangles. Using the Pythagorean theorem, we can calculate the distance as \(r = \sqrt{(\frac{a}{2})^2+(\frac{a}{2})^2+(\frac{a}{2})^2}\), which simplifies to \(r = \frac{a}{2}\sqrt{3}\). The electric field due to a single charge at the center of the cube is then \[E = \frac{kq}{(\frac{a}{2}\sqrt{3})^2} = \frac{kq}{\frac{3a^2}{4}}\]

Step 3: Calculate the total electric field

Due to the symmetry of the charges location, the magnitude of the electric field due to each charge at the cube's center is the same. We can divide the resulting electric fields into pairs, each pair pointing in opposite directions, the total electric field has a direction parallel to the direction of the electric field due to the single unpaired charge. Therefore, the total electric field at the center of the cube is \[E_{total} = E = \frac{4kq}{3a^2}\]

Step 4: Determine the answer

Now, we can substitute the value of Coulomb's constant (\(k = \frac{1}{4\pi\varepsilon_0}\)), \[E_{total} = \frac{4}{3a^2} \times \frac{1}{4\pi\varepsilon_0}q = \frac{1}{3\pi\varepsilon_0} \frac{q}{a^2}\] Comparing this with the given options, we find that our answer matches option (C). So, the magnitude of the electric field at the center of the cube is: \[\boxed{E_{total}= \frac{1}{3\pi\varepsilon_0} \frac{q}{a^2}}\]