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(B) The electric field $E$ due to an infinite charged cylinder at a distance $r$ from its axis is given by $E \propto \frac{1}{r}$.Since the $+Q$ char

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a force to the right and a clockwise torque

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(B) The electric field $E$ due to an infinite charged cylinder at a distance $r$ from its axis is given by $E \propto \frac{1}{r}$.Since the $+Q$ charge is closer to the cylinder than the $-Q$ charge,the electric field at the position of $+Q$ $(E_1)$ is stronger than the electric field at the position of $-Q$ $(E_2)$,i.e.,$E_1 > E_2$.The force on the $+Q$ charge is $F_1 = Q E_1$ (directed away from the cylinder,i.e.,to the right).The force on the $-Q$ charge is $F_2 = Q E_2$ (directed towards the cylinder,i.e.,to the left).Since $E_1 > E_2$,the net force $F_{	ext{net}} = F_1 - F_2$ is directed to the right.Regarding torque,the force $F_1$ acts at a greater distance from the center of the rod and is larger,while $F_2$ acts at a smaller distance. The combination of these forces creates a clockwise torque about the center of the rod.

Consider a positively charged infinite cylinder with uniform volume charge density. An electric dipole consisting of $+Q$ and $-Q$ charges attached to opposite ends of a massless rod is oriented as shown in the figure. At the instant as shown in the figure,the dipole will experience,

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