Two charges of $-4 \ \mu C$ and $+4 \ \mu C$ are placed at the points $A(1, 0, 4) \ m$ and $B(2, -1, 5) \ m$ located in an electric field $\vec{E} = 0.20 \ \hat{i} \ V/cm$. The magnitude of the torque acting on the dipole is $8 \sqrt{\alpha} \times 10^{-5} \ Nm$. Where $\alpha = $ . . . . . .

The force on a charge situated on the axis of a dipole is $F$. If the charge is shifted to double the distance,the new force will be

As shown in the figure,a particle $A$ of mass $2m$ and carrying charge $q$ is connected by a light rigid rod of length $L$ to another particle $B$ of mass $m$ and carrying charge $-q$. The system is placed in an electric field $\vec E$. The electric force on a charge $q$ in an electric field $\vec E$ is $\vec F = q \vec E$. After the system settles into equilibrium,one particle is given a small push in the transverse direction so that the rod makes a small angle $\theta_0$ with the electric field. Find the maximum tension in the rod.

Two opposite charges each of magnitude $500 \mu C$ are $10 \ cm$ apart. Find the electric field intensity at a distance of $25 \ cm$ from the midpoint on the axial line of the dipole.

If a dipole of dipole moment $\vec{p}$ is placed in a uniform electric field $\vec{E}$,then the torque acting on it is given by:

Two electric dipoles of moment $P$ and $8P$ are placed in opposite directions on a line at a distance of $24 \text{ cm}$. The electric field will be zero at a point between the dipoles whose distance from the dipole of moment $P$ is $x_1$. When $8P$ is replaced by $27P$ keeping all other quantities the same, the distance $x_1$ becomes $x_2$. The difference $|x_2 - x_1|$ is (All distances are measured from the centers of the dipoles.) (in $\text{ cm}$)