# Force-neutralizing charge (electrostatics)

## Homework Statement

A charge 2Q is placed a distance 'd' from charge 4Q. A third charge 'q' is placed 3m from charge 2Q directly on the line between charges 2Q and 4Q. Find d and q such that the force between charges 2Q and 4Q is equal to 0.

## Homework Equations

Coulomb's law: f = k(|q1q2|)/r2

k = 8.99 x 10^9 (Nm^2/c^2)

## The Attempt at a Solution

(k)(2Q)(q)/3 = [k(4Q)(q)]/(d - 3)

(d - 3)/3 = 2

d - 3 = 6;

d = 9m?

k(2Q)(q)/3 + [k(4Q)(q)]/6 = k(2Q)(4Q)/9;

q[(k4Q)/6 + (k4Q)/6] = k8Q2/9;

q[k4Q/3] = k8Q2/9;

q = (3)(2Q)/9 ?

q = 2Q/3 ?

cnh1995
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k(|q1q2|)/r2
(k)(2Q)(q)/3 = [k(4Q)(q)]/(d - 3)

Thanks for catching that.

(d - 3)2/3 = 2;

(d2 - 6d + 9)/3 = 2

d2 - 6d + 9 = 6

d2 - 6d + 3 = 0

-b +- (b2 - 4a)1/2/2a

(6 +- (24))1/2/2

d = 30 or d = -18?

cnh1995
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(d - 3)2/3 = 2;
There should be 9 in the denominator instead of 3.

Thanks, this isn't going so well for me :) :

(d - 3)2/9 = 2;

(d2 - 6d + 9)/9 = 2

d2 - 6d + 9 = 18

d2 - 6d - 9 = 0

I actually made a mistake for the quadratic formula part too in my last post:
-b +- (b2 - 4a)1/2 / 2a

d = [6 +- (24 - 4)1/2]/2

d = [6 +- (4.472)]/2

d = 5.23 or d = 0.764 ?

cnh1995
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b2 - 4a)1/2
It's (b2-4ac)1/2.
You'll get only one positive answer. Be careful with the signs and square terms.

Still not going well for me then..

-b +- (b2 - 4ac)1/2 / 2a

d = [6 +- (24 - (4)(1)(-9))1/2]/2

d = 6 +- (24 + 36)1/2 / 2

d = 6 +- (60)1/2/2

d = (6+- 7.745)/2

d = 6.8725 or d = -0.8725 ?

I'm going to assume use only the positive answer for d ? :

[k(2Q)(q)]/9 + [k(4Q)(q)]/(6.8725 - 3)2 = [k(2Q)(4Q)]/81;

[k(2Q)(q)]/9 + [k(4Q)(q)]/15= [k(2Q)(4Q)]/81;

q[(k4Q)/15 + (k2Q)/9] = (k8Q2)/81

q[(3)(k)(4Q) + (5)(k)(2Q)]/45 = (k8Q2)/81

q[(12)(k)(Q) + (10)(k)(Q)]/45 = (k8Q2)/81

q[k22Q/45] = (k8Q2)/81

q = [(45)(k8Q2)]/[(81)(k22Q)];

q = [360kQ2 /(1782kQ) ;

q = 0.202Q ?

cnh1995
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Still not going well for me then..

-b +- (b2 - 4ac)1/2 / 2a

d = [6 +- (24 - (4)(1)(-9))1/2]/2

d = 6 +- (24 + 36)1/2 / 2

d = 6 +- (60)1/2/2

d = (6+- 7.745)/2

d = 6.8725 or d = -0.8725 ?

I'm going to assume use only the positive answer for d ? :

[k(2Q)(q)]/9 + [k(4Q)(q)]/(6.8725 - 3)2 = [k(2Q)(4Q)]/81;

[k(2Q)(q)]/9 + [k(4Q)(q)]/15= [k(2Q)(4Q)]/81;

q[(k4Q)/15 + (k2Q)/9] = (k8Q2)/81

q[(3)(k)(4Q) + (5)(k)(2Q)]/45 = (k8Q2)/81

q[(12)(k)(Q) + (10)(k)(Q)]/45 = (k8Q2)/81

q[k22Q/45] = (k8Q2)/81

q = [(45)(k8Q2)]/[(81)(k22Q)];

q = [360kQ2 /(1782kQ) ;

q = 0.202Q ?
b2=36 and not 24..

I forgot the '-b' at the beginning as well...

Hopefully everything is more neat now:

-b +- (b2 - 4ac)1/2 / 2a

d = [-6 +- (62 - (4)(1)(-9))]/(2)

d = [-6 +- (36 + 36)]/2

d = [-6 +- (72)]/2

d = 33 or d = -34?

Assuming using only the positive answer again:

[k(2Q)(q)]/32 + [k(4Q)(q)]/(33 - 3)2 = [k(2Q)(4Q)]/81;

[k(2Q)(q)]/9 + [k(4Q)(q)]/900 = [(k)(2Q)(4Q)]/81;

[(100)(k)(2Q)(q)]/900 + [(k)(4Q)(q)]/900 = [(k)(2Q)(4Q)]/81;

[200Qkq + 4Qkq]/900 = [(k)(2Q)(4Q)]/81;

204Qkq/900 = [(k)(2Q)(4Q)]/81; now let's simplify the right hand side:

204Qkq/900 = [(k)(8Q2)]/81

q = [(900)(8Q2)(k)]/[(204)(81)(Q)(k)]

q = (7200Q)/(16524)

q = 0.435Q

cnh1995
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d = [-6 +- (72)**1/2**]/2
d = 33 or d = -34?
How come the test charge is only 3m away from 2Q but 30m away from 4Q?
(d - 3)2/9 = 2;
In the end, put your value of d in this expression and see if it satisfies the equation.

Last edited:
-b +- (b2 - 4ac)1/2 / 2a

d = [-6 +- (62 - (4)(1)(-9))1/2]/(2)

d = [-6 +- (36 + 36)1/2]/2

d = [-6 +- 8.4852]/2

d = 2.485 or d = -14.48 ?

(k)(2Q)(q)/32 + (k)(4Q)(q)/(2.485 - 3)2 = (k)(2Q)(4Q)/(2.485 + 3)2;

(k)(2Q)(q)/9 + (k)(4Q)(q)/0.265225 = (k)(8Q2)/30;

q[(137.72)(k)(Q)]/9 = (k)(8Q2)/30;

q = [(k)(9)(8Q2)]/[(30)(137.72)(k)(Q);

q = 0.01742Q

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cnh1995
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d = [-6 +- 8.4852]/2
-b=6 and not -6. It will give you
d=(6+8.4852)/2=7.2426 m.
You can verify the answer by putting it in this equation.
(d - 3)2/9 = 2;

Apologies for bumping this late I will try to be more active now. So in the beginning I had (corrected as) this:

2Q*q*k/32 = 4Q*q*k/(d - 3)2;

(d - 3)2/9 = 2;

And then I ended up later with this polynomial:

d2 - 6d - 9 = 0;

-B +- (B2 - 4ac)1/2/2a;

where a = 1, b = -6, and c = -9

d = 6 +- (36 + 36)1/2/2;

d = 6 +- (72)2/2;

d = 6 +- (8.4852)/2;

d = 7.2426 or d = -1.2426; and verifying it: (7.2426 - 3)2/9 = 2 -> 17.999/9 = almost 2, which would check out.

So what do I do about q now again? I get this doing it over again:

d = 7.2426;

(k)(2Q)(q)/32 + (k)(4Q)(q)/(7.2426 - 3)2 = (k)(2Q)(4Q)/(7.2426)2 which i corrected as it is d that was just the entire distance across;

Now factoring out q I get this:

q[k(2Q)/9 + k(4Q)/18] = k(8Q2)/52.4552 ;

q[k(4Q)/18 + k(4Q)/18] = k(8Q2)/52.4552 ;

q[k(8Q)/18] = k(8Q2)/52.4552 ;

q[k(8Q)/18] = k(8Q2)/52.4552;

q = [18 / 52.4552]*[k(8Q2)/k(8Q)] ;

q = [18 / 52.4552]Q;

q = 0.3431Q ?

haruspex
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I'm not sure I understand the question, and I certainly do not understand the attempted solution.
What is meant by "the force between charges 2Q and 4Q"? The obvious meaning is the force they exert on each other, but that would be unaffected by q. You seem to have interpreted it as the net force exerted on q, but it cannot be that since you would have no way to determine q.
How do you arrive at the equation here:
(k)(2Q)(q)/32+ (k)(4Q)(q)/(7.2426 - 3)2= (k)(2Q)(4Q)/(7.2426)2
You have the two forces exerted by 2Q and 4Q on q adding instead of opposing, and why should they add to the force 2Q and 4Q exert on each other?
A possible interpretation is that there should be no net force on either 2Q or 4Q. That would give you two equations with two unknowns. But if that is what the question means the wording is very poor.

A possible interpretation is that there should be no net force on either 2Q or 4Q.
Yes, this.

You have the two forces exerted by 2Q and 4Q on q adding instead of opposing, and why should they add to the force 2Q and 4Q exert on each other?
.
I made it that way because it was across the whole distance, but I probably should have made '(k)(2Q)(4Q)/(7.2426)2' negative, would this make more sense?

(k)(2Q)(q)/32 + (k)(4Q)(q)/(7.2426 - 3)2= -(k)(2Q)(4Q)/(7.2426)2

haruspex
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Yes, this.

I made it that way because it was across the whole distance, but I probably should have made '(k)(2Q)(4Q)/(7.2426)2' negative, would this make more sense?

(k)(2Q)(q)/32 + (k)(4Q)(q)/(7.2426 - 3)2= -(k)(2Q)(4Q)/(7.2426)2
Your equations still do not make any sense. Start again from scratch: what are the forces acting on the 2Q?
(You can drop the factors of k; clearly they will always cancel out.)

Okay, I think I know what you mean now. I really might need a refresher on basic mechanics (and everything, really):

Do I sum up the forces then, on each charge, like so?:

k(2Q)(q)/32 + k(4Q)(2Q)/d2 = k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2

haruspex
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Okay, I think I know what you mean now. I really might need a refresher on basic mechanics (and everything, really):

Do I sum up the forces then, on each charge, like so?:

k(2Q)(q)/32 + k(4Q)(2Q)/d2 = k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2
That's better, but what you have written is that the net force on the 2Q is equal to the net force on the 4Q. According to the interpretation you confirmed in your post #15, we know a bit more than that.

Okay so the forces on each charge are in opposite directions and on one side will be negative then, like this?:

k(2Q)(q)/32 + k(4Q)(2Q)/d2 = -[ k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2 ]

haruspex
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Okay so the forces on each charge are in opposite directions and on one side will be negative then, like this?:

k(2Q)(q)/32 + k(4Q)(2Q)/d2 = -[ k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2 ]
No, you misunderstand.
As you agreed, the condition is that the net force on each of 2Q and 4Q (separately) is zero.
So just consider the net force on one at a time. Don't try to combine them into one equation.

ahhh okay

so for 2Q I get this: k(2Q)(q)/32 + k(4Q)(2Q)/d2 = 0

and for 4Q I get this: k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2 = 0

My attempted work for solving for d now is as follows (where I will be solving 'q first' and then substituting back):

Using the 4Q equation:

k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2 = 0;

k(2Q)/d2 = -k(q)/(d - 3)2 ;

q = (-2Q)(d2 - 6d + 9)/d2 for 4Q;

Now for 2Q:

k(2Q)(q)/32 + k(4Q)(2Q)/d2 = 0;

k(2Q)(q)/32 = -k(4Q)(2Q)/d2 ;

q = (-9)(4Q)/d2 for 2Q; now substituting back:

(-9)(4Q)/d2 = (-2Q)(d2 - 6d + 9)/d2

-36Q = (-2Q)(d2 - 6d + 9) ;

-36Q = -2Qd2 + 12Qd - 18Q ;

0 = -2Qd2 + 12Qd + 18Q

Using: [-b +- (B2 - 4ac)1/2]/2a where a = -2Q, B = 12Q, C = 18Q

d = [-12Q +- (144Q2 - (4)(-2Q)(18Q))1/2]/-4Q

d = [-12Q +- (144Q2 + 144Q2)1/2]/-4Q ;

d = [-12Q +- 16.97Q]/-4Q

d = -28.9705/-4Q

d = 7.24264 or d = -1.2425 ??

cnh1995
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ahhh okay

so for 2Q I get this: k(2Q)(q)/32 + k(4Q)(2Q)/d2 = 0

and for 4Q I get this: k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2 = 0

My attempted work for solving for d now is as follows (where I will be solving 'q first' and then substituting back):

Using the 4Q equation:

k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2 = 0;

k(2Q)/d2 = -k(q)/(d - 3)2 ;

q = (-2Q)(d2 - 6d + 9)/d2 for 4Q;

Now for 2Q:

k(2Q)(q)/32 + k(4Q)(2Q)/d2 = 0;

k(2Q)(q)/32 = -k(4Q)(2Q)/d2 ;

q = (-9)(4Q)/d2 for 2Q; now substituting back:

(-9)(4Q)/d2 = (-2Q)(d2 - 6d + 9)/d2

-36Q = (-2Q)(d2 - 6d + 9) ;

-36Q = -2Qd2 + 12Qd - 18Q ;

0 = -2Qd2 + 12Qd + 18Q

Using: [-b +- (B2 - 4ac)1/2]/2a where a = -2Q, B = 12Q, C = 18Q

d = [-12Q +- (144Q2 - (4)(-2Q)(18Q))1/2]/-4Q

d = [-12Q +- (144Q2 + 144Q2)1/2]/-4Q ;

d = [-12Q +- 16.97Q]/-4Q

d = -28.9705/-4Q

d = 7.24264 or d = -1.2425 ??
This is no different from what you were doing earlier in the beginning of this thread where you got d=7.2426m. For cancellation of the forces,
i) Force on 2Q by q=Force on 2Q by 4Q
ii) Force on 4Q by q= force on 4Q by 2Q.
(with directions opposite to each other of course, since q is negative.)
Out of these, the right hand sides are equal, giving force on 2Q by q= force on 4Q by q. This is the equation you'd been working on in the beginning.

Quite an epic story about d. In reality, everything is not so grave)
Once, you get to the point
$$(d-3)^2=18$$
It means that
$$d-3 =\pm 3 \sqrt{2}$$
$$d =3(1 \pm \sqrt{2})$$
Certainly, you choose the positive answer for the distance (in meters)
$$d = 3 (1+ \sqrt{2})$$
Seems, that
$$|q|=\left ( \frac{2}{1+\sqrt{2}} \right )^2|Q|$$
Guess the sign?

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TSny
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Quite an epic story about d. In reality, everything is not so grave)