Question

Suppose the second charge in the previous problem is −1.0 × 10−6 C. Locate the position where a third charge will not experience a net force.

Answer 1

Answer:

1.0 × 10−6 C. can you please tell what you mean by the question

Answer 2

Explanation:

Step 1:

Given data in the question  

$q_{1}=2 \times 10^{-6} \mathrm{C}$

$q_{2}=-1 \times 10^{-6} C$  

Since both charges are opposite in nature, it is impossible to place the third charge between them. Let the third charge, q, be x cm away from load q1,

Step 2:

By the Law of Coulomb force;

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1} Q_{2}}{r^{2}}$

So, force q on charge because of q1,

$F=\frac{9 \times 10^{9} \times 2.0 \times 10^{-6} \times q}{x^{2}}$  

Force on charge q attributable to q2,

$F^{\prime}=\frac{9 \times 10^{9} \times 10^{-6} \times q}{(10-x)^{2}}$

Step 3:

As to the question,

$F-F^{\prime}=0$

$F=F^{\prime}$

$\frac{9 \times 10^{9} \times 2.0 \times 10^{-6} \times q}{x^{2}}=\frac{9 \times 10^{9} \times 10^{-6} \times q}{(10-x)^{2}}$

$x^{2}=2(10-x)^{2}$

$x^{2}-40 x+200=0$

$x=20 \pm 10^{2} \sqrt{2} m$

$x=34.14 \mathrm{cm}(\because x \neq 20-10 \sqrt{2})$