Question

What is the net electric force on charge B in FIGURE?

Short answer

The total electric force on charge,$F_{\text{net}}=-31.5\times {10}^{-5}\mathrm{N}$

Step-by-step solution

Charge on A,B and C

When two charged particles, one with charge $q_1$and also the other with charge $q_2$, are separated by a distance $r$, they exert forces on one other, with the force decreasing because the magnitude of the cost increases. This force is described by Coulomb's law, which is provided by equation.

$$\begin{gathered} F_{\mathrm{C}}=F_{1\text{on}2}=F_{2\text{on}1} \\ =\frac{K\left|q_1\right|\left|q_2\right|}{r^2} \end{gathered}$$

$K$is that the electrostatic constant, which is capable $9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2$. $nC$determines the charge of every sphere. by converting the unit to coulomb.

$$\begin{gathered} q_A=(+1\mathrm{nC})\left(\frac{1\times {10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right) \\ =+1\times {10}^{-9}\mathrm{C} \end{gathered}$$

$$\begin{gathered} q_B=(-2\mathrm{nC})\left(\frac{1\times {10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right) \\ =-2\times {10}^{-9}\mathrm{C} \end{gathered}$$

$$\begin{gathered} q_C=(+2\mathrm{nC})\left(\frac{1\times {10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right) \\ =+2\times {10}^{-9}\mathrm{C} \end{gathered}$$

The magnitude of the electrical force of A on B

Each of $A$and $C$'s spheres has an attraction force on $B$. Now we use the numbers for $q_A$, $q_B$, $r_{AB}$, and $K$to calculate the magnitude of electrical force on $A$and $B$.

$F_{A\text{on}B}=\frac{K{q}_A{q}_B}{r_{AB}^2}$

$=\frac{\left(9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(+1\times {10}^{-9}\mathrm{C}\right)\left(-2\times {10}^{-9}\mathrm{C}\right)}{(0.02\mathrm{m}{)}^2}$

$=-4.5\times {10}^{-5}\mathrm{N}$

Find the whole force

We also use the numbers for $q_C$, $q_B$, $r_{AB}$, and $K$to calculate the magnitude of $C$'s electric force on $B$.

$F_{ConB}=\frac{K{q}_C{q}_B}{r_{AB}^2}$

$=\frac{\left(9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(+2\times {10}^{-9}\mathrm{C}\right)\left(-2\times {10}^{-9}\mathrm{C}\right)}{(0.01\mathrm{m}{)}^2}$

$=-36\times {10}^{-5}\mathrm{N}$

due to two spheres $A$and $C$, the force on $B$is

$F_{\text{net}}=F_{C\text{on}B}-F_{A\text{on}B}$

$=-36\times {10}^{-5}\mathrm{N}-\left(-4.5\times {10}^{-5}\mathrm{N}\right)$

$=-31.5\times {10}^{-5}\mathrm{N}$

Because the magnitude of the force on the sphere $C$is greater than that on the sphere $A$, the force is directed forward.