Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 16, Problem 22Q

Given two point charges, Q and 2Q, a distance l apart, is there a point along the straight line that passes through them where E = 0 when their signs are (a) opposite, (b) the same? If yes, state roughly where this point will be.

Expert Solution
Check Mark
To determine

Part(a)To Determine:

A point along the straight line that passes through two charges where E=0 when the signs of the charges are opposite.

Answer to Problem 22Q

Solution:

The point along the straight line joining the two charges where E=0lies outside of the charges and at a distance of 2.41l away from the weaker charge and 3.41l away from the stronger charge if the charges have opposite signs and placed l distance apart.

Explanation of Solution

Given Info:

The two charges Q and 2Q are placed l distance apart.

Formula Used:

The electric field at a point r distance apart from a point charge q is given by,

E=kqr2

Calculation:

Let Q be negative and 2Q be positive. Since the two charges are of opposite signs then the point at which E=0 lies outside of the charges.

Physics: Principles with Applications, Chapter 16, Problem 22Q , additional homework tip  1

The electric fields at the point P due to the two charges are in opposite direction.

The point P is at a distance of x from the charge Q.

If E=0 at P, then EQ=E2Q at the point P

That is, kQx2=k×2Q(l+x)2

Therefore, 2x2=(l+x)2 and x22lxl2=0

Solving we get, x=2.41l

The point P lies at 2.41l away from Q and 3.41l away from 2Q

If two charges having opposite sign are placed at a distance l apart, then the point along the straight line joining the two charges where E=0 lies outside of the charges and at a distance of 2.41l away from the weaker charge and 3.41l away from the stronger charge.

Expert Solution
Check Mark
To determine

Part(b)To Determine:

A point along the straight line that passes through two charges where E=0 when the signs of the charges are the same.

Answer to Problem 22Q

Solution:

The point along the straight line joining the two charges where E=0 lies in between the charges and at a distance of 0.41l away from the weaker charge and 0.59l away from the stronger charge if the charges have same sign and placed l distance apart.

Explanation of Solution

Given Info:

The two charges Q and 2Q are placed l distance apart.

Formula Used:

The electric field at a point r distance apart from a point charge q is given by,

E=kqr2

Calculation:

Let both Q and 2Q are positive. Since the two charges are of same sign then

the point at which E=0 lies in between the charges.

Physics: Principles with Applications, Chapter 16, Problem 22Q , additional homework tip  2

The electric fields at the point P due to the two charges are in opposite direction.

The point P is at a distance of x from the charge Q.

If E=0

at P, then EQ=E2Q at the point P

That is, kQx2=k×2Q(lx)2

Therefore, 2x2=(lx)2 and x2+2lxl2=0

Solving we get, x=0.41l

The point P lies at 0.41l away from Q and 0.59l away from 2Q

If two charges having same sign are placed at a distance l apart, then the point along the straight line joining the two charges where E=0 lies in between the charges and at a distance of 0.41l away from the weaker charge and 0.59l away from the stronger charge.

Chapter 16 Solutions

Physics: Principles with Applications

Ch. 16 - Prob. 11QCh. 16 - Prob. 12QCh. 16 - Prob. 13QCh. 16 - Prob. 14QCh. 16 - Prob. 15QCh. 16 - Prob. 16QCh. 16 - Prob. 17QCh. 16 - Assume that the two opposite charges in Fig....Ch. 16 - Consider the electric field at the three points...Ch. 16 - Why can electric field lines never cross?Ch. 16 - Show, using the three rules for field lines given...Ch. 16 - Given two point charges, Q and 2Q, a distance l...Ch. 16 - Consider a small positive test charge located on...Ch. 16 - A point charge is surrounded by a spherical...Ch. 16 - Prob. 1PCh. 16 - Prob. 2PCh. 16 - Prob. 3PCh. 16 - Prob. 4PCh. 16 - Prob. 5PCh. 16 - Prob. 6PCh. 16 - Prob. 7PCh. 16 - Prob. 8PCh. 16 - Prob. 9PCh. 16 - Prob. 10PCh. 16 - Prob. 11PCh. 16 - Prob. 12PCh. 16 - Prob. 13PCh. 16 - Prob. 14PCh. 16 - Prob. 15PCh. 16 - Prob. 16PCh. 16 - Prob. 17PCh. 16 - Prob. 18PCh. 16 - Prob. 19PCh. 16 - Prob. 20PCh. 16 - Prob. 21PCh. 16 - Prob. 22PCh. 16 - Prob. 23PCh. 16 - Prob. 24PCh. 16 - Prob. 25PCh. 16 - Prob. 26PCh. 16 - Prob. 27PCh. 16 - Prob. 28PCh. 16 - Prob. 29PCh. 16 - Prob. 30PCh. 16 - Prob. 31PCh. 16 - Prob. 32PCh. 16 - Prob. 33PCh. 16 - Prob. 34PCh. 16 - Prob. 35PCh. 16 - Prob. 36PCh. 16 - Prob. 37PCh. 16 - Prob. 38PCh. 16 - Prob. 39PCh. 16 - Prob. 40PCh. 16 - Prob. 41PCh. 16 - Prob. 42PCh. 16 - Prob. 43PCh. 16 - Prob. 44PCh. 16 - Prob. 45PCh. 16 - Prob. 46PCh. 16 - Prob. 47PCh. 16 - Prob. 48PCh. 16 - Prob. 49PCh. 16 - Prob. 50PCh. 16 - Prob. 51PCh. 16 - Prob. 52GPCh. 16 - Prob. 53GPCh. 16 - Prob. 54GPCh. 16 - Prob. 55GPCh. 16 - Prob. 56GPCh. 16 - Prob. 57GPCh. 16 - Prob. 58GPCh. 16 - Prob. 59GPCh. 16 - Prob. 60GPCh. 16 - Prob. 61GPCh. 16 - Prob. 62GPCh. 16 - Prob. 63GPCh. 16 - Prob. 64GPCh. 16 - Prob. 65GPCh. 16 - Prob. 66GPCh. 16 - Prob. 67GPCh. 16 - Prob. 68GPCh. 16 - Prob. 69GPCh. 16 - Prob. 70GPCh. 16 - Prob. 71GPCh. 16 - Prob. 72GPCh. 16 - Prob. 73GP
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