91影视

A pair of equal and opposite charges separated by a distance is called a dipole.

If two charges have the same nature (Like charges) then they will repel each other or the force exerted on one charge due to the other charge would be repulsive in nature.

If two charges have opposite natures (Unlike charges) then they will attract each other or the force exerted on one charge due to the other charge would be attractive in nature.

Consider the given data as below.

The charges, $q_1=q_2=5.00渭C=5.00脳{10}^{鈭�6}\mathrm{C}$

The charge, ${\mathrm{q}}_3=10.00渭\mathrm{C}={\mathrm{q}}_3=10.00脳{10}^{鈭�6}\mathrm{C}$

The distance between two same charges, $\mathrm{d}=3.0\mathrm{cm}=0.03\mathrm{m}$

The distance between two charges, ${\mathrm{L}}_1={\mathrm{L}}_2={\mathrm{L}}_{\mathrm{L}}=2.0\mathrm{cm}=0.02\mathrm{m}$

Force between either charge of dipole and q₃ are equal but opposite in the direction.

The force on charge q₃ is given by

$\begin{array}{r}{\mathrm{F}}_{\text{net}}={\mathrm{F}}_1\sin 鈦�胃+{\mathrm{F}}_1\sin 鈦�胃\\ =2{\mathrm{F}}_1\sin 鈦�胃\end{array}$

The angle is given by

$\begin{array}{r}胃={\sin }^{鈭�1}鈦�\left(0.75\right)\\ ={48.6}^{鈭�}\end{array}$

Thus, the magnitude of force by coulomb鈥檚 law is

$F_1=\frac{k{q}_1{q}_3}{L^2}$

Substitute known values in the above equation, and you have

$\begin{array}{r}{F}_1=\frac{\left(9脳{10}^9\right)\left(5脳{10}^{鈭�6}\right)\left(10脳{10}^{鈭�6}\right)}{(0.02{)}^2}\\ =1.125脳{10}^3\mathrm{N}\end{array}$

${\mathrm{F}}_{\text{net}}=2{\mathrm{F}}_1\sin 鈦�胃$

Substitute known values in the above equation, and you have

role="math" localid="1668233461906" $\begin{array}{r}{\mathrm{F}}_{\text{net}}=2\left(1.125脳{10}^3\mathrm{N}\right)\sin 鈦�\left({48.6}^{鈭�}\right)\\ =1688\mathrm{N}\end{array}$

Therefore, the magnitude of force on ${\mathrm{q}}_3$is 1688 N and its direction is from (=5) to (-5) .

Electric field in the region is

$E=\frac{F}{q_3}$

Substitute known values in the above equation.

$\begin{array}{r}E=\frac{1088}{10脳{10}^{鈭�6}}\\ =1.68脳{10}^8\mathrm{N}/\mathrm{C}\end{array}$

The direction of E is downward q₃ charge tends to move in upward direction due to which q₂ tends to move in up and q₁ tends to move down, thus creating a counterclockwise torque of magnitude

$\begin{array}{r}\mathrm{蟿}={\mathrm{F}}_1\mathrm{dcos}鈦�\mathrm{胃}\\ =\left(1.125脳{10}^3\mathrm{N}\right)\left(0.03\mathrm{m}\right)\cos 鈦�\left(48.6\right)\\ =22.3\mathrm{N}鈰�\mathrm{m}\end{array}$

Therefore the magnitude of torque is role="math" localid="1668233695028" $22.3\mathrm{N}鈰�\mathrm{m}$and it is acting in counter-clockwise direction.