Question

charge Q1 is given to one plate of parallel plate capacitor and of capacitance c and Q2 charges given to other plate the potential difference developed between the two plate​

Answer 1

Given :

Charge given for first capacitor =Q₁

Charge given for second capacitor =Q₂

To Find :

The potential difference  between the two plates

Solution :

• Potential difference (V) = E×d

           V = E×d

• The net electric field is Enet = σ₁ - σ₂ / 2ε₀

          V = (σ₁ - σ₂ / 2ε₀) × d

• σ is surface charge density which is stated as charge per unit area

           V = (Q₁/A - Q₂/A)×d / 2ε₀

           V = (Q₁ - Q₂ ) / (2ε₀A / d)

• The capacitance between two plates is given by  C =Aε₀ / d

           V =  (Q₁ - Q₂ ) / C

The potential difference between two plates is   (Q₁ - Q₂ ) / C

Answer 2

Explanation:

(Q1-Q2)/2C

sigma is the surface charge density