Question

a charge q is placed at the centre of the line joining two equal changes Q such that the system is in equilibrium calculated the value of q in term of Q​

Answer 1

Answer:

Given:

2 charges Q are separated by a distance (say r ) . Another charge q is placed at centre of the separation and system remains in equilibrium.

To find:

q in terms of Q

Concept:

We can clearly understand that q has to be a negative charge for system to be in equilibrium . For example :

If we consider a terminal Q , it will be pulled by q and pushed by another terminal Q. Only then it can remain in equilibrium.

Calculation:

The Pushing and Pulling forces will be equal :

$\sf{F_{1} = F_{2}}$

$\sf{ \implies \dfrac{1}{4\pi \epsilon} \dfrac{Qq}{ { (\frac{r}{2}) }^{2} } = \dfrac{1}{4\pi \epsilon} \dfrac{QQ}{ { r }^{2} } }$

$\sf{ \implies 4q = Q }$

$\sf{ \implies q = \dfrac{Q}{4} }$

Inserting the sign of charge q :

$\sf{ \implies q = ( - \dfrac{ Q}{4} })$

So final answer :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \large{ \sf{ \red{ \bold{ q = ( - \dfrac{ Q}{4} })}}}}$

Answer 2

$\huge{ \underline{ \rm{Given:-}}}$

2 charges Q are separated by a distance (say r ) . Another charge q is placed at centre of the separation and system remains in equilibrium.

$\huge{ \underline{ \rm{To \: find:-}}}$

q in terms of Q.

$\huge{ \underline{ \rm{Calculation:-}}}$

Pushing and pulling forces be equal, so :

F1= F2

$\implies \: \frac{1}{4 \pi \: e} \frac{Qq}{( \frac{r}{2}) ^{2} } = \frac{1}{4 \pi \: e} \frac{QQ}{r {}^{2} } \\ \implies \: 4q = Q \:$

Inserting the sign of charge q, we get :

$\implies \: q = (\frac { - Q}{4} )$

$\huge{ \red{ \boxed{ \underline{ \rm{Answer:- }}}}}$

$\implies \: q = (\frac { - Q}{4} )$