The image contains a diagram and a question related to the concept of electric fields and charges with reference to Gauss's Law and the principle of superposition.### Diagram Explanation:- **Plates with Charge:** - Two parallel plates are depicted. - The left plate has a negative charge density, represented as "-σ." - The right plate has a positive charge density, represented as "σ = Q/A." - **Electric Field Lines:** - Between the plates, electric field lines are shown. - The field between the plates is noted as "E = σ/ε₀" directed from the positive plate towards the negative plate. - Outside the left plate, the electric field is "σ/2ε₀" directed to the left. - Outside the right plate, the electric field is "σ/2ε₀" directed to the left.### Question:The question asks about the net electric field to the right of the positive plate and the reason for this according to the principles taught in class.- **Options:** 1. \( \frac{\sigma}{\varepsilon_0} \), Gauss's law 2. \( \frac{\sigma}{\varepsilon_0} \), superposition 3. 0, superposition 4. \( \frac{\sigma}{2\varepsilon_0} \), superpositionStudents are instructed to use "+" to denote directions to the right and "-" to denote directions to the left, with "Q" referring to a positive charge.

Answered: 6 E=2€₁ 26₁ 6= D=0 LE=E₁ + + ola A As…

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6 E=2€₁ 26₁ 6= D=0 LE=E₁ + + ola A As described in class, the net electric field to the right of the positive plate is_ of because Use + to denote directions to the right and to denote directions to the left. Assume the symbol Q refers to a positive charge.