Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1634

Two air capacitors \(A=1 \mu F, B=4 \mu F\) are connected in series with $35 \mathrm{~V}\( source. When a medium of dielectric constant \)\mathrm{K}=3$ is introduced between the plates of \(\mathrm{A}\), change on the capacitor changes by (A) \(16 \mu \mathrm{c}\) (B) \(32 \mu \mathrm{c}\) (C) \(28 \mu \mathrm{c}\) (D) \(60 \mu \mathrm{c}\)

Short Answer

Expert verified
The change in the charge on the capacitor is \(\Delta Q = 32 \mu C\). The correct answer is (B) \(32 \mu C\).

Step by step solution

01

Determine the equivalent capacitance before introducing the dielectric

The given capacitors, A and B, are connected in series. To find the equivalent capacitance of the series connection, use the formula: \(\frac{1}{C_{eq}} = \frac{1}{C_A} + \frac{1}{C_B}\) where \(C_{eq}\) is the equivalent capacitance, \(C_A = 1 \mu F\) is the capacitance of capacitor A, and \(C_B = 4 \mu F\) is the capacitance of capacitor B. Substitute the values and find \(C_{eq}\): \(\frac{1}{C_{eq}} = \frac{1}{1 \mu F} + \frac{1}{4 \mu F}\) Solving for the equivalent capacitance, we get: \(C_{eq} = 0.8 \mu F\)
02

Calculate the charge on capacitors A and B before introducing the dielectric

To find the charge on capacitors A and B, use the formula: \(Q_{before} = C_{eq}V\) where \(Q_{before}\) is the charge before introducing the dielectric, and \(V = 35V\) is the voltage source. Substitute the values and find \(Q_{before}\): \(Q_{before} = 0.8 \mu F \cdot 35V\) Solving for the charge, we get: \(Q_{before} = 28 \mu C\)
03

Determine the new capacitance of capacitor A after introducing the dielectric

When a dielectric medium with dielectric constant \(K = 3\) is introduced to capacitor A, its capacitance changes as follows: \(C_{A'} = KC_A\) where \(C_{A'}\) is the new capacitance of capacitor A. Substitute the values and find \(C_{A'}\): \(C_{A'} = 3 \cdot 1 \mu F\) Solving for the new capacitance, we get: \(C_{A'} = 3 \mu F\)
04

Determine the new equivalent capacitance and charge after introducing the dielectric

Using the formula for capacitors connected in series as in Step 1, we now have \(C_A = 3 \mu F\) and \(C_B = 4 \mu F\). Calculate the new equivalent capacitance: \(\frac{1}{C_{eq'}} = \frac{1}{3 \mu F} + \frac{1}{4 \mu F}\) Solving for the new equivalent capacitance, we get: \(C_{eq'} = 1.714 \mu F\) Now, calculate the new charge on the capacitors: \(Q_{after} = C_{eq'}V\) Substitute the values and find \(Q_{after}\): \(Q_{after} = 1.714 \mu F \cdot 35V\) Solving for the charge, we get: \(Q_{after} = 60 \mu C\)
05

Calculate the change in charge

To find the change in charge, subtract the initial charge from the final charge: \(\Delta Q = Q_{after} - Q_{before}\) Substitute the values and find the change in charge: \(\Delta Q = 60 \mu C - 28 \mu C\) Solving for the change in charge, we get: \(\Delta Q = 32 \mu C\) Therefore, the correct answer is (B) \(32 \mu C\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)

Two small conducting sphere of equal radius have charges \(+1 \mathrm{c}\) and \(-2 \mathrm{c}\) respectively and placed at a distance \(\mathrm{d}\) from each other experience force \(F_{1}\). If they are brought in contact and separated to the same distance, they experience force \(F_{2}\). The ratio of \(F_{1}\) to \(F_{2}\) is \(\ldots \ldots \ldots \ldots\) (A) \(-8: 1\) (B) \(1: 2\) (C) \(1: 8\) (D) \(-2: 1\)

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) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) / \mathrm{c}\right]\)

A thin spherical shell of radius \(R\) has charge \(Q\) spread uniformly over its surface. Which of the following graphs, figure most closely represents the electric field \(\mathrm{E}\) (r) produced by the shell in the range $0 \leq \mathrm{r}<\infty\(, where \)\mathrm{r}$ is the distance from the centre of the shel1.

Two thin wire rings each having a radius \(R\) are placed at a distance \(d\) apart with their axes coinciding. The charges on the two rings are \(+q\) and \(-q\). The potential difference between the centers of the two rings is $\ldots .$ (A) 0 (B) $\left.\left[\mathrm{q} /\left(2 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{(}^{2}+\mathrm{d}^{2}\right)\right\\}\right]$ (C) $\left[\mathrm{q} /\left(4 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{\left. \left.\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)\right\\}\right]}\right.\right.$ (D) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)
See all solutions

Recommended explanations on English Textbooks

Language Acquisition

Read Explanation

Lexis and Semantics Summary

Read Explanation

Linguistic Terms

Read Explanation

Syntax

Read Explanation

Essay Plan

Read Explanation

Global English

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free