Problem 88 - A Complete Resource Book in Physics for JEE Main

Question

A long string with a charge of \(\lambda\) per unit length passes through an imaginary cube of edge \(a\). The maximum flux of the electric field through the cube will be (A) \(\lambda a / \varepsilon_{0}\) (B) \(\frac{\sqrt{2} \lambda a}{\varepsilon_{0}}\) (C) \(\frac{6 \lambda a^{2}}{\varepsilon_{0}}\) (D) \(\frac{\sqrt{3} \lambda a}{\varepsilon_{0}}\)

Step-by-Step Solution

Answer

The maximum electric flux through the imaginary cube is (C) \(\frac{6 \lambda a^{2}}{\varepsilon_{0}}\).

Step 1: Consider one face of the cube

We first start by considering one face of the cube. Take the cube to have its center at the origin and the long string to pass through one of the edges of the cube. The center of the face of the cube closest to the string will have the maximum electric field. This is also true for the other faces.

Step 2: Calculate the electric field intensity

Using Gauss Law, we know that the electric field intensity for a long straight wire is given by the formula: \[E = \frac{\lambda}{2\pi\varepsilon_{0}r}\] where \(E\) is the electric field intensity, \(\lambda\) is the charge per unit length, \(\varepsilon_{0}\) is the permittivity of free space, and \(r\) is the distance from the wire. In this case, as the string passes through the edge, the distance from the center of the nearest face to the string is \(r = a/2\). Plug this value into the formula: \[E = \frac{\lambda}{2\pi\varepsilon_{0}\cdot(a/2)}\]

Step 3: Calculate the flux through one face of the cube

Now that we have the electric field intensity, we can find the electric flux through the face. To do that, we need to find the component of the electric field that is perpendicular to the face: \[E_\bot = E \cdot \cos{\theta}\] For a cube, the angle between the electric field and the area vector of the face is 45 degrees, so \(\cos{45^{\circ}} = \frac{1}{\sqrt{2}}\): \[E_\bot = \frac{\lambda}{2\pi\varepsilon_{0}\cdot(a/2)} \cdot \frac{1}{\sqrt{2}}\] Finally, to find the flux through one face, we multiply the perpendicular electric field by the area of the face, which is \(a^2\): \[\Phi_1 = E_\bot \cdot a^2 = \frac{\lambda a}{2\pi\varepsilon_{0}\cdot(a/2)} \cdot \frac{1}{\sqrt{2}} \cdot a^2\]

Step 4: Calculate the total flux through all faces of the cube

There are 6 faces in total on the cube. Multiply the electric flux calculated in the previous step by 6 to find the total electric flux through all faces of the cube: \[\Phi = 6\cdot \Phi_1 = 6\cdot \frac{\lambda a}{2\pi\varepsilon_{0}\cdot(a/2)} \cdot \frac{1}{\sqrt{2}} \cdot a^2\] Simplify the expression: \[\Phi = \frac{6\lambda a}{\varepsilon_{0}}\] Hence, the maximum electric flux through the cube is: \[\Phi = \frac{6\lambda a}{\varepsilon_{0}}\] Therefore, the correct option is (C) \(\frac{6 \lambda a^{2}}{\varepsilon_{0}}\).