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

A Charge \(q\) is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel is $\ldots \ldots \ldots \ldots$ (A) \(\left(\mathrm{q} / \in_{0}\right)\) (B) (q / \(2 \in_{0}\) ) (C) \(\left(2 q / \epsilon_{0}\right)\) (D) Zero

The inward and outward electric flux for a closed surface in units of \(\mathrm{Nm}^{2} / \mathrm{C}\) are respectively \(8 \times 10^{3}\) and $4 \times 10^{3}\(. Then the total charge inside the surface is \)\ldots \ldots \ldots \ldots . . \mathrm{c}$. (A) \(\left[\left(-4 \times 10^{3}\right) / \epsilon_{0}\right]\) (B) \(-4 \times 10^{3}\) (C) \(4 \times 10^{3}\) (D) \(-4 \times 10^{3} \mathrm{E}_{0}\)

Two charged spheres of radii \(R_{1}\) and \(R_{2}\) having equal surface charge density. The ratio of their potential is ..... (A) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)\) (B) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)^{2}\) (C) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}\) (D) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)\)

Two point charges \(-q\) and \(+q\) are located at points \((0,0,-a)\) and $(0,0, a)\( respectively. The potential at a point \)(0,0, z)\( where \)z>a\( is \)\ldots \ldots$ (A) $\left[(2 \mathrm{q} a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}+a^{2}\right)\right\\}\right]$ (B) \(\left[\mathrm{q} /\left(4 \pi \epsilon_{0} \mathrm{a}\right)\right]\) (C) \(\left[\right.\) (qa) \(\left./\left(4 \pi \in_{0} z^{2}\right)\right]\) (D) $\left[(2 q a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}-a^{2}\right)\right\\}\right]$

An oil drop of 12 excess electrons is held stationary under a constant electric field of \(2.55 \times 10^{4} \mathrm{Vm}^{-1}\). If the density of the oil is \(1.26 \mathrm{gm} / \mathrm{cm}^{3}\) then the radius of the drop is \(\ldots \ldots \ldots \mathrm{m}\). (A) \(9.75 \times 10^{-7}\) (B) \(9.29 \times 10^{-7}\) (C) \(9.38 \times 10^{-8}\) (D) \(9.34 \times 10^{-8}\)
See all solutions

Recommended explanations on English Textbooks

Morphology

Read Explanation

Creative Story

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Phonetics

Read Explanation

Email

Read Explanation

Cues and Conventions

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