Question

Two large, thin, parallel sheets have surface charge densities of opposite signs and equal magnitude \(\sigma\). What is the magnitude of the electric field \((E)\) in the region between the sheets?

• A. \(\dfrac{\sigma}{2\varepsilon_{0}}\)
• B. \(\dfrac{\sigma}{\varepsilon_{0}}\)
• C. Zero
• D. \(\dfrac{2\sigma}{\varepsilon_{0}}\)

Hint:

Remember the electric field due to an infinite charged sheet: \[ \boxed{E=\frac{\sigma}{2\varepsilon_{0}}} \] For two parallel sheets having equal and opposite charges, \[ \boxed{ E_{\text{between}} = \frac{\sigma}{\varepsilon_{0}} } \] while \[ \boxed{ E_{\text{outside}}=0. } \]

Answer 1

The Correct Option is B

Concept: According to Gauss's Law, the electric field due to an infinite plane sheet carrying a uniform surface charge density \(\sigma\) is \[ \boxed{E=\frac{\sigma}{2\varepsilon_{0}}} \] This electric field is independent of the distance from the sheet. For two parallel sheets carrying equal and opposite surface charge densities, \[ +\sigma \qquad \text{and} \qquad -\sigma, \] the electric fields combine according to the principle of superposition.

• Between the sheets, both electric fields act in the same direction and hence add.

• Outside the sheets, the electric fields are equal in magnitude but opposite in direction and therefore cancel each other.

Step 1: Electric field due to a positively charged sheet.
For the positively charged sheet, \[ E_{+}=\frac{\sigma}{2\varepsilon_{0}}. \] The field is directed away from the positive sheet.

Step 2: Electric field due to a negatively charged sheet.
For the negatively charged sheet, \[ E_{-}=\frac{\sigma}{2\varepsilon_{0}}. \] The field is directed towards the negative sheet.

Step 3: Find the resultant electric field between the sheets.
Between the sheets, both electric fields are in the same direction. Hence, \[ E=E_{+}+E_{-}. \] Substituting, \[ E = \frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}}. \] Therefore, \[ E = \frac{\sigma}{\varepsilon_{0}}. \] Thus, \[ \boxed{E=\frac{\sigma}{\varepsilon_{0}}.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (B)}}. \]