Question

Three infinitely long charged thin sheets are placed as shown in the figure. The magnitude of the electric field at the point \( P \) is \(\frac{x \sigma}{\epsilon_0}\). The value of \( x \) is _________.
(All quantities are measured in SI units).

Answer 1

Correct Answer: 2

The electric field at \( P \) is the vector sum of the fields due to the three charged sheets.

Using Gauss’s Law:

\[ \vec{E} = \frac{\sigma}{2\epsilon_0} \]

For the point \( P \):

\[ \vec{E}_P = \left( \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} \right) (-\hat{i}) \]

Simplify:

\[ \vec{E}_P = \frac{4\sigma}{2\epsilon_0} (-\hat{i}) = \frac{2\sigma}{\epsilon_0} (-\hat{i}) \]

Thus, \( x = 2 \).

Answer 2

The problem requires us to find the magnitude of the net electric field at a point P due to three infinitely long charged thin sheets. The sheets are placed at positions \(X = -a\), \(X = a\), and \(X = 3a\) with surface charge densities \(-\sigma\), \(-2\sigma\), and \(\sigma\) respectively. The net electric field at P is given in the form \(\frac{x\sigma}{\epsilon_0}\), and we need to determine the value of \(x\).

Concept Used:

The electric field produced by an infinite, non-conducting thin sheet of charge with a uniform surface charge density \(\rho\) is uniform and perpendicular to the sheet. Its magnitude is given by:

\[ E = \frac{|\rho|}{2\epsilon_0} \]

where \(\epsilon_0\) is the permittivity of free space. The direction of the electric field is away from the sheet for a positive charge density and towards the sheet for a negative charge density.

The total electric field at any point due to multiple charged sheets is the vector sum of the electric fields produced by each individual sheet. This is known as the principle of superposition.

Step-by-Step Solution:

Step 1: Identify the charge densities and positions of the three sheets.

• Sheet 1: at \(X = -a\), with charge density \(\sigma_1 = -\sigma\).
• Sheet 2: at \(X = a\), with charge density \(\sigma_2 = -2\sigma\).
• Sheet 3: at \(X = 3a\), with charge density \(\sigma_3 = \sigma\).

The point P is located between the sheets at \(X=a\) and \(X=3a\).

Step 2: Determine the electric field vector due to each sheet at point P. Let \(\hat{i}\) be the unit vector in the positive X-direction.

The electric field due to Sheet 1 (\(\sigma_1 = -\sigma\)) at point P: Since P is to the right of this sheet and the charge is negative, the field points towards the sheet, i.e., in the \((-\hat{i})\) direction.

\[ \vec{E}_1 = \frac{|-\sigma|}{2\epsilon_0}(-\hat{i}) = -\frac{\sigma}{2\epsilon_0}\hat{i} \]

The electric field due to Sheet 2 (\(\sigma_2 = -2\sigma\)) at point P: Since P is to the right of this sheet and the charge is negative, the field points towards the sheet, i.e., in the \((-\hat{i})\) direction.

\[ \vec{E}_2 = \frac{|-2\sigma|}{2\epsilon_0}(-\hat{i}) = -\frac{2\sigma}{2\epsilon_0}\hat{i} = -\frac{\sigma}{\epsilon_0}\hat{i} \]

The electric field due to Sheet 3 (\(\sigma_3 = \sigma\)) at point P: Since P is to the left of this sheet and the charge is positive, the field points away from the sheet, i.e., in the \((-\hat{i})\) direction.

\[ \vec{E}_3 = \frac{|\sigma|}{2\epsilon_0}(-\hat{i}) = -\frac{\sigma}{2\epsilon_0}\hat{i} \]

Step 3: Apply the superposition principle to find the net electric field at P.

The net electric field \(\vec{E}_P\) is the vector sum of the individual fields:

\[ \vec{E}_P = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 \] \[ \vec{E}_P = \left(-\frac{\sigma}{2\epsilon_0}\hat{i}\right) + \left(-\frac{\sigma}{\epsilon_0}\hat{i}\right) + \left(-\frac{\sigma}{2\epsilon_0}\hat{i}\right) \]

Step 4: Simplify the expression for the net electric field.

\[ \vec{E}_P = \left(-\frac{\sigma}{2\epsilon_0} - \frac{2\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0}\right)\hat{i} \] \[ \vec{E}_P = \left(-\frac{\sigma + 2\sigma + \sigma}{2\epsilon_0}\right)\hat{i} = -\frac{4\sigma}{2\epsilon_0}\hat{i} \] \[ \vec{E}_P = -\frac{2\sigma}{\epsilon_0}\hat{i} \]

Final Computation & Result:

The magnitude of the net electric field at point P is:

\[ E_P = |\vec{E}_P| = \left|-\frac{2\sigma}{\epsilon_0}\hat{i}\right| = \frac{2\sigma}{\epsilon_0} \]

We are given that the magnitude of the electric field is \(\frac{x\sigma}{\epsilon_0}\). Comparing this with our result:

\[ \frac{x\sigma}{\epsilon_0} = \frac{2\sigma}{\epsilon_0} \]

By comparing the two expressions, we find that:

The value of x is 2.