Question

If two charges +q and +4q are seprated by distance d and a third charge q is placed on the line joining the above two charges such that all are in equilibrium. The magnitude, sign and position of charge q are:

Answer 1

Answer:

$\large\bold\red{\frac{4q}{9}\;C\:;\:+ve\:or\:-ve\:nature\:\:;\:\frac{2d}{3}\:from\:+4q}$

Explanation:

Let,

The charge +q is placed at point M

And,

The charge +4q is placed at point N

It's given that,

Separation between both the charges is d

Therefore,

Distance between M and N = d.

Now,

Let a third charge Q is placed at r distance from N.

Therefore,

Distance from M will be (d-r).

Now,

From Coulomb's Law,

Force on charge Q due to +q = $\bold{\frac{kQq}{ {(d - r)}^{2} } }$

And,

Force on charge Q due to +4q = $\bold{\frac{4kQq}{ {(r)}^{2} } }$

Now,

Since the charges are in equilibrium,

Therefore,

We get,

$= > \frac{kQq}{ {(d - r)}^{2} } = \frac{4kQq}{ {r}^{2} } \\ \\ = > 4 {(d - r)}^{2} = {r}^{2} \\ \\ = > 2(d - r) = r \\ \\ = > 2d = 3r \\ \\ = > \bold{r = \frac{2d}{3} }$

Therefore,

The charge Q is at a distance of $\bold{\frac{2d}{3}}$ from charge +4q.

Now,

To check the magnitude of the charge Q,

Since charges are in equilibrium,

Therefore,

Net force due to +q and Q on +4q will be equal.

Therefore,

We get,

$= > \frac{4kQq}{ {d}^{2} } = \frac{4kqq}{ {( \frac{d}{3} )}^{2} } \\ \\ = > Q {d}^{2} = \frac{4q {d}^{2} }{9} \\ \\ = >\large\bold{ Q = \frac{4q}{9}\:C}$

Hence,

The charge Q is positive or negative both.

Answer 2

Answer:

Therefore, charge $Q$ exists placed at a distance $d / 3$ from the charge $q$ and it exists negative in sign and magnitude is $4 \mathrm{q} / 9$.

Explanation:

Let the Magnitude of the third charge be $Q$ and It exists placed at a distance $r$ from the point $A$ where the charge $\mathrm{q}$exists a place. So the distance of charge $\mathrm{Q}$ from the charge $+4q$ will be $(d-r)$.

Let the force on the charge $+\mathrm{q}$ due to the third charge $Q$ exists $F_{13}$ and Force on the charge $+4 \mathrm{q}$ due to the third charge $\mathrm{Q}$ is $\mathrm{F}_{23}$.

Now by using Coulomb's Law,

We have,

$F_{13}=k \frac{Q q}{r^{2}}$

and

$F_{23}=k \frac{4 Q q}{(d-r)^{2}}$

Since it exists given that system exists beneath equilibrium conditions therefore these two forces must exist equivalent and in opposing directions. Therefore, the third $Q$ must be negative in the sign

So we have,

$&F_{13}=F_{23}$

$&k \frac{Q q}{r^{2}}=k \frac{4 Q r}{(d-r)^{2}}$

$&\frac{1}{r^{2}}=\frac{4}{(d-r)^{2}}$

equating, we get

$&4 r^{2}=(d-r)^{2}$

$&2 r=d-r$

$&2 r+r=d$

$&3 r=d$

then $&r=\frac{d}{3}\end{aligned}$

Now,

Let us use the equilibrium state for a charge $+q$

We have,

$&k \frac{Q q}{\left(\frac{d}{3}\right)^{2}}=k \frac{4 q-q}{d^{2}}$

$&\frac{9 Q}{d^{2}}=\frac{4 q}{d^{2}}$

$&Q=\frac{4 q}{9}\end{aligned}$

Therefore charge $Q$ exists placed at a distance $d / 3$ from the charge $q$ and it exists negative in sign and magnitude is $4 \mathrm{q} / 9$.

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