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Chapter 13: Problem 113

Total charge on plate (2) initially is (A) \(\frac{\varepsilon_{0} A}{2 d} V\) (B) \(\frac{2 \varepsilon_{0} A}{d} V\) (C) \(\frac{\varepsilon_{0} A}{d} V\) (D) \(\frac{3 \varepsilon_{0} A}{2 d} V\)

Short Answer

Expert verified
The total charge on plate (2) initially can be found using the formula \(Q = CV\), where C is the capacitance of a parallel plate capacitor, given by \(\frac{\varepsilon_{0} A}{d}\). Substituting this capacitance formula into the charge formula, we get \(Q = \frac{\varepsilon_{0} A}{d} V\). Therefore, the correct answer is (C) \(\frac{\varepsilon_{0} A}{d} V\).

Step by step solution

01

Understand the relationship between charge, capacitance, and voltage

To calculate the total charge on a plate, we need to use the basic formula relating capacitance (C), voltage (V), and charge (Q): \[Q = C V\]
02

Calculate capacitance of a parallel plate capacitor

The capacitance of a parallel plate capacitor can be found using the following equation: \[C = \frac{\varepsilon_{0} A}{d}\] where \(\varepsilon_{0}\) is the permittivity of free space, A is the area of the plate, and d is the distance between the plates.
03

Substitute capacitance formula to calculate Q

Now plug in the formula for the capacitance of a parallel plate capacitor into the charge formula: \[Q = C V = \left(\frac{\varepsilon_{0} A}{d}\right)V\]
04

Simplify the expression

Simplify the expression: \[Q = \frac{\varepsilon_{0} A}{d} V\]
05

Find the correct option

Comparing our calculated expression with the given options, we can see that the correct answer is: (C) \(\frac{\varepsilon_{0} A}{d} V\)

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Most popular questions from this chapter

A point charge \(q\) and a charge \(-q\) are placed at \(x=-a\) and \(x=+a\), respectively. Which of the following represents a part of \(E-x\) graph?

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