Question

44. What is the force $\overrightarrow{F}$ on the $-1.0nC$ charge in the middle of FIGURE P$22.44$due to the four other charges? Give your answer in component form.

Short answer

As a result, the net force on the charge in the center is $\left(1.0\times {10}^{-3}\mathrm{N}\right)\hat{\mathbf{i}}$.

Step-by-step solution

Step 1:Given data

The general equation for the force between two charged particles is as follows:,

$F_{\mathrm{e}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d^2}$

Here ${\epsilon }_0$is the vacuum permittivity, and d is the distance between the point charges

Step 2:Explain diagram

Consider the diagram below.

Step 3:Net force

The shown direction indicates the directions of electric force at the center of the square. Because 2 and 3 are positive charges, they attract the charge, whereas 1 and 4 are negative charges. They repel the charge away from them in the same force direction as 2 and 3. Because all four forces are of equal magnitude, the resultant force will be in the positive X direction, as indicated by the arrow. The force will not have a Y component. It will only have one X component that points in the positive X direction..

Substituting $1.0\mathrm{nC}$for $q_1,2.0\mathrm{nC}$for $q_2,9\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2$for $\frac{1}{4\pi {\epsilon }_0}$and$\frac{1}{\sqrt{2}}\mathrm{cm}$for d in the expression of electrostatic force and solve,

$F_e=\frac{\left(9\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\left(1.0\mathrm{nC}\right)\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)\right)\left(\left(2.0\mathrm{nC}\right)\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{{\left(\left(\frac{1}{\sqrt{2}}\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}$

$=3.6\times {10}^{-4}\mathrm{N}$

Now, the net force acting on a charged particle is given by the horizontal component of all four forces, which is calculated as follows:

$F_{\mathrm{net}}=4F_e\cos \left(\frac{\pi }{4}\right)$

Substitute $3.6\times {10}^{-4}\mathrm{N}$for $F_e$in above equation.

$F_{\text{net}}=4\left(3.6\times {10}^{-4}\mathrm{N}\right)\cos \left(\frac{\pi }{4}\right)$

$=1.0\times {10}^{-3}\mathrm{N}$

The net force is in the X direction..

$F_{\text{net}}=\left(1.0\times {10}^{-3}\mathrm{N}\right)\widehat{i}$

As a result, the net force on the charge in the center is$\left(1.0\times {10}^{-3}\mathrm{N}\right)\hat{\mathbf{i}}$.