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Two identical metal plates are given positive charges q_1 and q_2 respectively where q_1<q_2 If they are brought close together to form a parallel plate with capacitance C, then the potential difference between them is:

Option: 1

q_1+q_2 / 2 C


Option: 2

q_1+q_2 / C


Option: 3

q_1-q_2 / C


Option: 4

\left(q_1-q_2\right) /(2 \mathrm{C})


Answers (1)

best_answer

The potential difference between the two identical metal plates is given as 

                                                    \mathbf{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}

Let the surface charge density is given as 

                                                   \sigma=\frac{\mathrm{q}}{\mathrm{A}}

The net electric field is 

                                       \mathrm{E}_{\mathrm{net}}=\frac{\sigma_1-\sigma_2}{2 \varepsilon_0}

We know the potential difference is given as 

                                                     \mathrm{V}=\mathrm{E} \cdot \mathrm{d}

By substituting the above values we get 

                                                       \mathrm{V}=\frac{\mathrm{q}_1-\mathrm{q}_2}{2 \mathrm{C}}

Posted by

Gaurav

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