ElectroMagnetic Question?

Question

Heya i am having trouble in answering this question

A negative charge –Q is located between two positive charges, +4Q and +16Q. The distance between +4Q and –Q is d1 and the distance between +16Q and –Q is d2.

All charges lie along the same line and d1+d2 = 9 cm.

Calculate d1 and d2 which will provide a stable equilibrium position for

the charge –Q, i.e., the location where the total force acting on –Q due

to +4Q and +16Q is zero.

Answer 1

hey dude i will help u in this

let +4Q charge is at point A AND +16Q charge is at B then  

Electric field due to A is =1/kxQa

                           Ea=9.09x109X4/d12

Eb=9.09X109X16/d22

But net force on –Q charge is 0 which is only possible when it is placed in b/w

These two charges i.e. net-Q=Ea-Eb=0

This mean              Ea=Eb

i.e. 9.09X109X4/d1=9.09X109X16/d22

4/d12=16/d22

4d22=16d12  

2d1=4d2

d1=2d2

d1+d2=9cm

d2=9-d1

d1=2(9-d1)

d1=18-2d1

3d1=18

d1=6cm and d2=3cm

  

it is right

Answer 2

ask dr. armstrong, or dr. smith, from Voltex V. They always use electromagnetic energy to beat the Bozenians and defend the earth lol.

Answer 3

Do you understand the question and not know how to answer it, or do you not understand the question?   I believe it is easier not to use the charge values at all.

Let's simplify this to...

***Knowns***

Equation 1: Qo = -Q

Equation 2: Q1 = +4Q

Equation 3: Q2 = +16Q = 4Q1

Equation 4: d1 + d2 = 9cm, which can be written d1 = 9cm - d2

The really helpful relationship we'll find by doing this symbolicly is Q2 = 4Q1.  This ratio will be crucial in solving the problem.  You can get the same ratio by using the charge values, but I think it's more straightforward to do so symbolically from the start.

Now, Coulomb's Law states that the force on a charge due to another charge is k * (the two charges) / (distance separating them)^2.  In this case, we know the magnitude of the force on Qo due to Q1 is equal to the magnitude of the force due to Q2 on Qo (because it's in equilibrium between the two).  Let's call the force due to Q1 F1, and the force due to Q2 F2.

F1 = F2

which is

kQoQ1 / d1^2 = kQoQ2 / d2^2

First, you see the kQo can be divided from both sides.

Q1 / d1^2 = Q2 / d2^2

From here, it's simple algebra.  We can eliminate the variables using the known equations.  In particular, after we plug in equations 3 an 4, it becomes...

Q1 / (9cm - d2)^2 = 4Q1 / (d2)^2

Now you can divide both sides by Q1...

1 / (9cm - d2)^2 = 4 / (d2)^2

And just solve for d2.  Then to get d1, use d1 = 9cm - d2.

Answer 4

I'm sure you would find the answer in your text books..or among your peers. Good luck!