Question

Two point charges +16 µC and -9 µC are placed 8 cm apart in air. You are asked to place +10 µC at a third position such that the net force on the +10 µC charge is zero. Where will you place the charge? Make the necessary calculations.

Answer 1

Answer:

24cm from -9 micro C charge

wish my explanation is helpful to you

Answer 2

Answer:

Two charges 16µC and -9 µC are placed at a separation of 8m, a third charge is placed such that, it experiences no net force due to these charges, the third charge should place at a distance of $24cm$ from the first charge

Explanation:

• The electric force due to two point charges ($q_{1} ,q_{2}$) separated by a distance (r) is given by the formula

       $E=K\frac{q_{1} q_{2} }{r^{2} }$

       where the constant $K=9*10^{9}$

From the question, we have two charges arranged as below

Let the third charge  +10 µC be placed at a distance (x) from the second charge

Force on it due to $q_{1}$ is $F_{1} =\frac{Kq_{1} 10}{(8+x)^{2} }$

Force on it due to $q_{2}$ is $F_{2} =\frac{Kq_{2} 10}{x^{2} }$

no net force on the third charge means, the charge is at equilibrium

that is the force on (Q) due to the two charges are equal and opposite

⇒ $\frac{Kq_{1} 10}{(8+x)^{2} } =\frac{Kq_{2}10 }{x^{2} }$

⇒$\frac{K*16*10}{(8+x)^{2} } =\frac{K*9*10 }{x^{2} }$

⇒$\frac{9}{16} =\frac{x^{2} }{(8+x)^{2} }$

take square root

⇒$3/4=x/8+x$

cross multiply

⇒

$24+3x=4x\\24=4x-3x=x\\x=24cm$

the third charge should place at a distance of $24cm$ from the second charge