Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1585

Electric charges of \(+10 \mu \mathrm{c}, 5 \mu \mathrm{c},-3 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed at the corners of a square of side $\sqrt{2 m}\( the potential at the centre of the square is \)\ldots \ldots$ (A) \(1.8 \mathrm{~V}\) (B) \(1.8 \times 10^{5} \mathrm{~V}\) (C) \(1.8 \times 10^{6} \mathrm{~V}\) (D) \(1.8 \times 10^{4} \mathrm{~V}\)

Short Answer

Expert verified
The potential at the center of the square is (D) \(1.8 \times 10^{4} V\).

Step by step solution

01

Determine the distance to the center

Since the charges are on a square with side length \(\sqrt{2m}\), we need to find the distance of each charge to the center of the square. Since the square is symmetric, the distances for all the charges are the same. We can find the distance from one corner to the center by using the Pythagorean theorem: \(r = \sqrt{(\frac{\sqrt{2m}}{2})^2 + (\frac{\sqrt{2m}}{2})^2} = \sqrt{\frac{2m}{4}+\frac{2m}{4}} = \sqrt{\frac{2m}{2}} = \sqrt{m}\)
02

Calculate potential for each charge

Now we need to calculate the electric potential due to each charge at the center of the square. We will use the formula for electric potential, \(V = \frac{kQ}{r}\), and the distances calculated in step 1. Charge 1: \(V_1 = \frac{9 \times 10^9 (+10 \times 10^{-6})}{\sqrt{2}} \) Charge 2: \(V_2 = \frac{9 \times 10^9 (+5 \times 10^{-6})}{\sqrt{2}} \) Charge 3: \(V_3 = \frac{9 \times 10^9 (-3 \times 10^{-6})}{\sqrt{2}} \) Charge 4: \(V_4 = \frac{9 \times 10^9 (+8 \times 10^{-6})}{\sqrt{2}} \)
03

Sum the potentials

Since the electric potential is a scalar quantity, we can simply add the potentials from each charge to find the net electric potential at the center of the square. \(V_\text{net} = V_1 + V_2 + V_3 + V_4\) Substituting the expressions from step 2: \(V_\text{net} = \frac{9 \times 10^9}{\sqrt{2}}(10 \times 10^{-6} + 5 \times 10^{-6} - 3 \times 10^{-6} + 8 \times 10^{-6})\)
04

Simplify and find the answer

Now, let's simplify and calculate the net electric potential: \(V_\text{net} = \frac{9 \times 10^9}{\sqrt{2}}(20 \times 10^{-6})\) \(V_\text{net} = 1.8 \times 10^4 V\) So, the correct answer is (D) \(1.8 \times 10^{4} V\).

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

64 identical drops of mercury are charged simultaneously to the same potential of 10 volt. Assuming the drops to be spherical, if all the charged drops are made to combine to form one large drop, then its potential will be (A) \(100 \mathrm{~V}\) (B) \(320 \mathrm{~V}\) (C) \(640 \mathrm{~V}\) (D) \(160 \mathrm{~V}\)

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed \(10 \mathrm{~cm}\) apart in air. The work done to bring them $4 \mathrm{~cm}$ closer is (A) Zero (B) \(4.8 \mathrm{~J}\) (C) \(3.5 \mathrm{~J}\) (D) \(-5.8 \mathrm{~J}\)

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 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) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)

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) \(-4 \times 10^{3} \mathrm{E}_{0}\)
See all solutions

Recommended explanations on English Textbooks

Graphology

Read Explanation

Linguistic Terms

Read Explanation

Multiple Choice Questions

Read Explanation

Rhetoric

Read Explanation

Lexis and Semantics Summary

Read Explanation

Phonology

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