Question

Two fixed point charges $+16\mu C$ and $-9\mu C$ are placed $8cm$ apart in air. Where should a third point charge placed so that, the third charge is in equilibrium

• A. $0.24m$ from the charge $+16\mu c$ and within the system of charge
• B. $0.24m$ from the charge $+16\mu c$ and outside the system along the straight line joining the two charges
• C. $0.24m$ from the charge $-9\mu c$ and within the system of charges
• D. $0.24m$ from the $-9\mu c$ and outside the system along the straight lint joining the two charges.

Answer 1

Step 1: Given the data and diagram used

$0.24m$ from the charge $+16\mu c$ and within the system of charge

Step 2: Calculation of distance of where third point charge placed so that, the third charge is in equilibrium

$F_{-q}=\frac{K\left(-9\right)\times \theta }{{\left(0.08+x\right)}^2},F_{+16q}=\frac{K\times 15\mu C}{{\left(x\right)}^2}$

$\left|F_{-q}\right|=\left|F_{+16q}\right|\Rightarrow \frac{9}{\left(0.8+x^2\right)}=\frac{16}{x^2}$

${\left(\frac{x}{0.8+x}\right)}^2=\left(\frac{16}{9}\right)\Rightarrow \frac{x}{0.8+x}=\frac{4}{3}$

$\therefore \left[\right(16\left)/\right(x+0.08{)}^2]=(9/x2)$

$\therefore \left[x/\right(x+0.08){]}^2=(9/16\left)\right[x/(x+0.8)]$

$=\left(3/4\right)4x=3x+0.24x=0.24m=24cm$

24 cm from −9μC outside their joining line

$x=0.24m$

The correct option is A