Question

Two Identical charges ‘q’each are kept at a distance ‘r’from
each other. A third charge Q' is placed on the line joining the two
bodies such that all the three charges are in equilionum. What is
the magnitude and nature of third charge.

Answer 1

Explanation:

let us suppose that the third charge Q is placed on the line joining the first and the second charge such that AO= x and OB = r-x

Net force on each of the three charges are zero as the system is in equilibrium.

If we assume that charge q is taken as negative then,on charge Q the forces will act in opposite direction therefore third charge should be positive

Now using Coulombs law

KQq/x^2= KQq /(r-x)^2

x^2 = (r-x)^2

x=(r-x)

Hence x= r/2

Now to find the magnitude of the charge we have to consider the case where charge Q at A or at B are in equilibrium

i.e, KQq/(r/2)^2 = KQQ / r^2

therefore q= Q/4

Answer 2

✪ᴀɴsᴡᴇʀ✪

• Magnitude of third charge is $\frac{q}{4}$ and is opposite in sign to the given charges.

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

Let Q be third charge(sign included)

✯ʟᴏᴄᴀᴛɪᴏɴ ᴏғ ᴛʜɪʀᴅ ᴄʜᴀʀɢᴇ✯

Externally Placed

Let us suppose that Q is placed externally on the line of the given charges.In this case, both charges will either attract or repel the third charge and as such Q will not be in equilibrium.

Internally Placed

As placing Q in an external location disrupts equilibrium, Q must lie in between the two charges.Let Q is

• d distance far from first charge
• (r-d) distance far from second charge

Since Q must be in equilibrium, net force on Q must be zero.

$\vec{F}_{31}+\vec{F}_{32} = 0$

$⇒ \frac{KQq}{d²}\hat{r}_{31}+\frac{KQq}{(r-d)²}\hat{r}_{32}= 0$

Since $\hat{r}_{31}=-\hat{r}_{32}$

$⇒ \frac{KQq}{d²}\hat{r}_{31}-\frac{KQq}{(r-d)²}\hat{r}_{31}= 0$

$⇒ \frac{KQq}{d²}-\frac{KQq}{(r-d)²}= 0$

$⇒ \frac{KQq}{d²}=\frac{KQq}{(r-d)²}$

$⇒d²=(r-d)²$

$⇒d=±(r-d)$

$⇒d=r-d$ or $d=-r+d$

$⇒2d=r$ or $0=-r$

$⇒d=\frac{r}{2}$

Thus Q lies exactly halfway between the given charges.

✫ᴍᴀɢɴɪᴛᴜᴅᴇ ᴏғ ᴛʜɪʀᴅ ᴄʜᴀʀɢᴇ✫

Since all three charge must be in equilibrium, net force on the given charge 'q' must be zero.

$\vec{F}_{13}+\vec{F}_{12} = 0$

$⇒\frac{KQq}{(\frac{r}{2})²}\hat{r}_{13}\frac{Kqq}{r²}\hat{r}_{12}=0$

$⇒\frac{4KQq}{r²}\hat{r}_{12}+\frac{Kqq}{r²}\hat{r}_{12}=0$

$⇒\frac{4KQq}{r²} +\frac{Kqq}{r²}=0$

$⇒4Q+q=0$

$⇒4Q =-q$

$⇒Q =-\frac{q}{4}$

Thus Q is of magnitude $\frac{q}{4}$ and opposite in sign of q.