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Chapter 11: Problem 1630

Two identical metal plates are given positive charges \(\mathrm{Q}_{1}\) and \(\mathrm{Q}_{2}\left(<\mathrm{Q}_{1}\right)\) respectively. If they are now brought close to gather to form a parallel plate capacitor with capacitance \(\mathrm{c}\), the potential difference between them is (A) \(\left[\left(Q_{1}+Q_{2}\right) /(2 c)\right]\) (B) \(\left[\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right) / \mathrm{c}\right]\) (C) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) /(2 \mathrm{c})\right]\) (D) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) / \mathrm{c}\right]\)

Short Answer

Expert verified
The short answer to the question is: The potential difference between the two metal plates is \(V = \frac{(Q_1 + Q_2)}{2C}\), which matches with option (A).

Step by step solution

01

Identifying the formula for capacitance and potential difference

First, we need to recall the formula for capacitance of a parallel plate capacitor and the relation between potential difference, charge, and capacitance. The capacitance formula for a parallel plate capacitor is \(C = \frac{Q}{V}\) where C is the capacitance, Q is the charge on the plate, and V is the potential difference between the plates. We need to find V, so we can rearrange the formula to: \(\displaystyle V=\frac{Q}{C}\) Now, we have to find the values of Q and C, keeping in mind that the charges on the plates are given as \(Q_1\) and \(Q_2\).
02

Calculate charge on the plates

When the two metal plates are brought close together to form a parallel plate capacitor, the charges on the plates will redistribute themselves such that the total charge on both plates remains constant. If the plates are identical, the charge on each plate will be the average of both charges. So, we can calculate the new charges on each plate: \(Q_{avg} = \frac{Q_1 + Q_2}{2}\) Now, since both plates are identical, let Q be the charge on each plate after redistribution: \(Q = Q_{avg}\)
03

Calculate the potential difference

Now that we have the charge on each plate (Q), we can use the rearranged capacitance formula from Step 1 to calculate the potential difference between the plates: \(\displaystyle V=\frac{Q}{C}\) Plugging in the values from Step 2: \(V = \frac{Q_{avg}}{C} = \frac{\frac{Q_1 + Q_2}{2}}{C}\)
04

Write the final answer and compare with the given options

The potential difference between the metal plates is: \(V = \frac{(Q_1 + Q_2)}{2C}\) Comparing our answer with the given options, we find that our answer matches with option (A): (A) \(\left[\left(Q_{1}+Q_{2}\right) /(2 c)\right]\) Hence, the correct choice is option (A).

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Most popular questions from this chapter

Electric potential at any point is $\mathrm{V}=-5 \mathrm{x}+3 \mathrm{y}+\sqrt{(15 \mathrm{z})}$, then the magnitude of the electric field is \(\ldots \ldots \ldots \mathrm{N} / \mathrm{C}\). (A) \(3 \sqrt{2}\) (B) \(4 \sqrt{2}\) (C) 7 (D) \(5 \sqrt{2}\)

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Equal charges \(q\) are placed at the vertices \(A\) and \(B\) of an equilateral triangle \(\mathrm{ABC}\) of side \(\mathrm{a}\). The magnitude of electric field at the point \(c\) is \(\ldots \ldots \ldots\) (A) \(\left(\mathrm{Kq} / \mathrm{a}^{2}\right)\) (B) \(\left.(\sqrt{3} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (C) \(\left.(\sqrt{2} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (D) $\left[\mathrm{q} /\left(2 \pi \mathrm{t} \varepsilon_{0} \mathrm{a}^{2}\right)\right]$
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