# Two charges of  3 C  and  -1 C are positioned on a line at points  6  and  5 , respectively. What is the net force on a charge of  2 C at  -2 ?

May 31, 2017

$\text{Net force on charge 2C :"4.7*10^8N" ,at negative x direction.}$

#### Explanation:

• The position of the point electric charges in the symbolic figure above is exaggerated.
• Coulomb's law states that electrical charges will force each other.
• It is a force pushing or pulling force between two charges.
• If the charges are the same, pushing is made, if it is reversed, pulling force occurs.
• We can calculate the force that the electric charges apply to each other by using the formula given below.

$F = K \cdot \frac{{q}_{1} \cdot {q}_{2}}{d} ^ 2$
$\text{where ;}$
${q}_{1} : \text{first charge}$
${q}_{2} : \text{second charge}$
$d : \text{distance between "q_1" and } {q}_{2}$
$k : {9.10}^{9} N \cdot {m}^{2} \cdot {C}^{- 2}$

• Both the B and C spheres apply force to the A sphere.
• The net force applied to A is equal to the vector sum of the forces.
• Let's calculate the force that B applies to A.

$\textcolor{b l u e}{{F}_{\text{BA}} = K \cdot \frac{- 1 \cdot 2}{7} ^ 2 = - \frac{2 K}{49}}$

• Let's calculate the force that C applies to A.

$\textcolor{g r e e n}{{F}_{\text{CA}} = K \cdot \frac{3 \cdot 2}{8} ^ 2 = \frac{6 K}{64}}$

• Now let's find the vector sum.

Sigma F_("net")=color(blue)(F_("BA"))+color(green)(F_("CA"))

$\Sigma {F}_{\text{net}} = \frac{- 2 K}{49} + \frac{6 K}{64}$

$\Sigma {F}_{\text{net}} = \frac{- 128 K + 294 K}{3136}$

$\Sigma {F}_{\text{net}} = \frac{166 K}{3136}$

$\Sigma {F}_{\text{net}} = \frac{166 \cdot 9 \cdot {10}^{9}}{3166}$

$\Sigma {F}_{\text{net}} = 0.47 \cdot {10}^{9}$

$\Sigma {F}_{\text{net}} = 4.7 \cdot {10}^{8} N$

• The net force is in negative x direction.