91影视

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

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

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

Consider the diagram below.

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 $q_2,9脳{10}^9\mathrm{N}路{\mathrm{m}}^2/{\mathrm{C}}^2$for $\frac{1}{4蟺{蔚}_0}$and$\frac{1}{\sqrt{2}}\mathrm{cm}$for d in the expression of electrostatic force and solve,

$F_e=\frac{\left(9脳{10}^9\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\left(1.0\mathrm{nC}\right)\left(\frac{{10}^{鈭�9}\mathrm{C}}{1\mathrm{nC}}\right)\right)\left(\left(2.0\mathrm{nC}\right)\left(\frac{{10}^{鈭�9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{{\left(\left(\frac{1}{\sqrt{2}}\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}$

$=3.6脳{10}^{鈭�4}\mathrm{N}$

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

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

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

$F_{\text{net}}=4\left(3.6脳{10}^{鈭�4}\mathrm{N}\right)\cos 鈦�\left(\frac{蟺}{4}\right)$

$=1.0脳{10}^{鈭�3}\mathrm{N}$

The net force is in the X direction..

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

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