21.55 • three charges are at the corners of an isosceles triangle as shown in fig. e21.55. the ±5.00 μc…

Explanation:

Step1: Recall Coulomb's law

The force between two point - charges is given by $F = k\frac{q_1q_2}{r^{2}}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them.

Step2: Calculate the force on each charge of the dipole due to the $- 10\ \mu C$ charge

Let $q_1=-10\times10^{- 6}\ C$, $q_{+}=5\times10^{-6}\ C$, $q_{-}=-5\times10^{-6}\ C$, and $r = 2\times10^{-2}\ m$. The force on the $+5\ \mu C$ charge due to the $-10\ \mu C$ charge is $F_{+}=k\frac{|q_1q_{+}|}{r^{2}}=(9\times10^{9})\frac{(10\times10^{-6})(5\times10^{-6})}{(2\times10^{-2})^{2}}=\frac{9\times10^{9}\times50\times10^{- 12}}{4\times10^{-4}} = 112.5\ N$. The force on the $-5\ \mu C$ charge due to the $-10\ \mu C$ charge is $F_{-}=k\frac{|q_1q_{-}|}{r^{2}}=(9\times10^{9})\frac{(10\times10^{-6})(5\times10^{-6})}{(2\times10^{-2})^{2}} = 112.5\ N$.

Step3: Find the net force on the dipole

The forces on the two charges of the dipole due to the $-10\ \mu C$ charge are equal in magnitude. The direction of the force on the $+5\ \mu C$ charge is towards the $-10\ \mu C$ charge and the direction of the force on the $-5\ \mu C$ charge is away from the $-10\ \mu C$ charge. The net force on the dipole is $F_{net}=F_{+}-F_{-}=0\ N$ (because of the symmetry of the isosceles - triangle arrangement).

Step4: Calculate the torque on the dipole

The dipole moment $p = qd$, where $q = 5\times10^{-6}\ C$ and $d$ is the distance between the two charges of the dipole. The distance between the $+5\ \mu C$ and $-5\ \mu C$ charges can be found using the Pythagorean theorem. If the height of the isosceles triangle is $h = 3\times10^{-2}\ m$ and the slant - side is $r = 2\times10^{-2}\ m$, the distance between the $+5\ \mu C$ and $-5\ \mu C$ charges $d=\sqrt{(2\times10^{-2})^{2}-(1.5\times10^{-2})^{2}}\times2=\sqrt{4\times10^{-4}-2.25\times10^{-4}}\times2=\sqrt{1.75\times10^{-4}}\times2\approx2.65\times10^{-2}\ m$. The dipole moment $p=(5\times10^{-6})\times(2.65\times10^{-2})\ C\cdot m$. The electric field due to the $-10\ \mu C$ charge at the mid - point of the dipole is $E = k\frac{|q_1|}{r_{mid}^{2}}$, where $r_{mid}$ is the distance from the $-10\ \mu C$ charge to the mid - point of the dipole. Using the Pythagorean theorem, $r_{mid}=\sqrt{(3\times10^{-2})^{2}+(1.325\times10^{-2})^{2}}\approx3.28\times10^{-2}\ m$. The torque on the dipole is $\tau = pE\sin\theta$. Since the axis is perpendicular to the dipole, $\theta = 90^{\circ}$ and $\sin\theta = 1$. Another way: The force on each charge of the dipole due to the $-10\ \mu C$ charge is $F = k\frac{|q_1q|}{r^{2}}=(9\times10^{9})\frac{(10\times10^{-6})(5\times10^{-6})}{(2\times10^{-2})^{2}} = 112.5\ N$. The distance between the lines of action of the two forces on the dipole charges is $d = 3\times10^{-2}\ m$. The torque $\tau = Fd=(112.5)\times(3\times10^{-2}) = 3.375\ N\cdot m$. The direction of the torque is such that it will tend to rotate the dipole clock - wise (using the right - hand rule).

Answer:

(a) Magnitude: $0\ N$, Direction: None (b) Magnitude: $3.375\ N\cdot m$, Direction: Clock - wise (using the right - hand rule)