A charge +Q is located at the origin and a second charge, +4Q, is at distance d on the x-axis. If a third charge, q, is placed somewhere, so that all three charges will be in equilibrium.1. What should be its sign, so that all three charges will be in equilibrium?A. Negative. B. Positive.2. What should be its magnitude, so that all three charges will be in equilibrium?

1). The sign that will be of all the given three charges being in equilibrium would be:

A). Negative

2). The magnitude to allow all the given three charges to remain in equilibrium would be:

- Magnitude

Given that,

First Charge

Second Charge

Third Charge

The location where they will be in equilibrium would be:

So,

We will assume the distance between the first two charges i.e. +Q and +4Q as d and assume the distance from charge +Q to the third charge q as x.

Now,

Therefore,

The equilibrium will be at:

by solving we get,

Because at equilibrium ,

So,

F

and

by resolving the equation, we get;

By solving it, we get

Thus, the sign is negative.  

and Magnitude |q| = 4Q/3

Thus, options A and would be the correct answer.

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Third charge should be placed at a distance d/3

Sign is negative

Magnitude is |q| = 4Q/3

Explanation:

To find a place where a third charge, q, is placed so that all three charges will be in equilibrium would be at;

E1 = E2

Now, let the distance between the charges +Q and +4Q be d. Also, let the distance from charge +Q to the third charge q be x.

Thus;

E1 = kQ/x²

E2 = k(4Q)/(d - x)²

Since equilibrium, then;

E1 = E2 gives;

kQ/x² = k(4Q)/(d - x)²

k and Q will cancel out to give;

1/x² = 4/(d - x)²

Cross multiply to get;

(d - x)² = 4x²

Taking square root of both sides gives;

√(d - x)² = √4x²

Gives;

d - x = 2x

d = x + 2x

d = 3x

x = d/3

This is the point at which E1 = E2.

Since at equilibrium, it means that Fq + F2 = 0

Now, electric force formula is;

F = kq1*q2/r²

So; Fq = kqQ/x²

F2 = kQ(4Q)/d²

Equating Fq + F2 = 0 gives;

kqQ/x² + kQ(4Q)/d² = 0

kqQ/x² = -kQ(4Q)/d²

Like terms cancel out to get;

q/x² = -4Q/d²

q = - 4Qx²/d²

earlier we saw that x = d/3.

So, q = -4Q × (d/3)²/d²

q = -4Q × ⅓

q = - 4Q/3

Thus,it's sign is negative.

It's magnitude will be the positive value.

Thus, magnitude is;

|q| = 4Q/3