Question

Two point charges q1 = +1μC and q2= +4 μ are placed 2m apart in air. At what distance from q1 along the line joining the two charges, will the net electric field be zero ? ​

Answer 1

Given :

Two-point charges $\sf Q_1=1\mu\ C$ and  $\sf Q_2=4\mu\ C$ are placed 2m apart in the air.

To be found :

Distance from $\sf Q_1$ along the line joining the two charges, will the net electric field be zero.

Theory :

• Electric field intensity :

Electric field intensity at a point due to a given charge is defined as the force experienced by a unit positive charge placed from that point.

$\purple{\boxed{ \vec{E}=\lim{q_o\to\ 0}\frac{ \vec{F}}{q_o}}}$

Solution :

Let point A where the charge $\sf\ Q_1$ is placed at B be where $\sf\ Q_2$

and O be is the point where the Electric field is zero.

Let r distance from $\sf Q_1$ along the line joining the two charges, will the net electric field be zero.

Thus,

AO = r m

and OB = 2-r m

Since $\sf \vec{E_{net}}=0$

$\sf \vec{E_{Q_1}}=\vec{E_{Q_2}}$

$\sf \implies \frac{kq_1}{r^{2}}=\frac{kQ_2}{(2-r)^2}$

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

On taking square root both sides,

$\sf \implies \frac{1}{r}=\frac{2}{2-r}$

$\sf \implies 2-r=2r$

$\sf \implies 2=3r$

$\sf \implies r=\frac{2}{3}$

Therefore, 2/3m Distance from $\sf Q_1$ along the line joining the two charges, will the net electric field be zero.