Question

Two particles A and B having charges 8×10^6C and -2×10^-6 respectively are held fixed at the separation of 20cm.Where should a third particle be placed so that it doesn't experience a net electric force?

Answer 1

it should be placed at a distance 0.2m from B

Answer 2

Answer:

For the third particle not to experience a net electric force it should be placed at a distance of $\bold{\pm 8.94 \times 10^{6}}$.

Explanation:

The two particles are placed at a fixed distance between them, then a third charge is introduced, its position to let the charge do not experience any net electric force that is the system be in equilibrium can be calculated as follows -  

Given, the charge q1 = $8 \times 10^{6}$C

Charge q2 = -$2 \times 10^{-6}$C

Charge q3 = q

Distance between q1 and q2 = 20  

Distance between q1 and q3 = 20 - a

Distance between q2 and q3 = a

Now, to have equilibrium of the system, force on charge q1 and q2 exerted by q3 must be equal, so, we have –

Force on q3  by q1 = force on q3 by q2

$\begin{aligned} \frac{k \times q_{3} \times q_{1}}{(20-a)^{2}} &=\frac{k q_{3} \times q_{2}}{a^{2}} \\ \frac{k \times 8 \times 10^{6}}{k \times 2 \times 10^{6}} &=\frac{(20-a)^{2}}{a^{2}} \end{aligned}$

$\begin{array}{l}{-4 \times 10^{12}=\frac{(20-\alpha)^{2}}{a^{2}}} \\ {-4 \times 10^{12} a^{2}=(20-a)^{2}} \\ {-4 \times 10^{12} a^{2}=400+a^{2}-2 a \times 20} \\ {400+a^{2}+4 \times 10^{12} a^{2}-40 a=0} \\ {400+5 \times 10^{12} a^{2}-40 a=0} \\ {80+10^{12} a^{2}-8 a=0}\end{array}$

Solving this equation, we get a = $\pm 8.94 \times 10^{6}$

So the charge q3 be placed at the distance of $\bold{\pm 8.94 \times 10^{6}}$ from the charge q2.