An electric dipole of moment $$p$$ is lying along a uniform electric field $$E.$$ The work done in rotating the dipole by $${90^ \circ }$$ is
A.
$$\sqrt 2 pE$$
B.
$$\frac{{pE}}{2}$$
C.
$$2pE$$
D.
$$pE$$
Answer :
$$pE$$
Solution :
When an electric dipole is placed in an electric field $$E,$$ a torque $$\tau = p \times E$$ acts on it. This torque tries to rotate the dipole.
If the dipole is rotated from an angle $${\theta _1}$$ to $${\theta _2}$$ then work done by external force is given by
$$W = pE\left( {\cos {\theta _1} - \cos {\theta _2}} \right)\,......\left( {\text{i}} \right)$$
Putting $${\theta _1} = {0^ \circ },{\theta _2} = {90^ \circ }$$ in the Eq. (i), we get
$$\eqalign{
& W = pE\left( {\cos {0^ \circ } - \cos {{90}^ \circ }} \right) \cr
& = pE\left( {1 - 0} \right) = pE \cr} $$
Releted MCQ Question on Electrostatics and Magnetism >> Electric Field
Releted Question 1
A hollow metal sphere of radius $$5 cms$$ is charged such that the potential on its surface is $$10\,volts.$$ The potential at the centre of the sphere is
A.
zero
B.
$$10\,volts$$
C.
same as at a point $$5 cms$$ away from the surface
D.
same as at a point $$25 cms$$ away from the surface
Two point charges $$ + q$$ and $$ - q$$ are held fixed at $$\left( { - d,o} \right)$$ and $$\left( {d,o} \right)$$ respectively of a $$x-y$$ coordinate system. Then
A.
The electric field $$E$$ at all points on the $$x$$-axis has the same direction
B.
Electric field at all points on $$y$$-axis is along $$x$$-axis
C.
Work has to be done in bringing a test charge from $$\infty $$ to the origin
D.
The dipole moment is $$2qd$$ along the $$x$$-axis
Three positive charges of equal value $$q$$ are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in
A uniform electric field pointing in positive $$x$$-direction exists in a region. Let $$A$$ be the origin, $$B$$ be the point on the $$x$$-axis at $$x = + 1cm$$ and $$C$$ be the point on the $$y$$-axis at $$y = + 1cm.$$ Then the potentials at the points $$A,B$$ and $$C$$ satisfy: