Question 31:
easyAn electric dipole is placed at an angle of $30^\circ$ with an electric field intensity $2 \times 10^5 \text{ N/C}$. It experiences a torque equal to $4 \text{ Nm}$. The charge on the dipole, if the dipole length is $2 \text{ cm}$, is: (2016-II)
Torque $\tau = pE\sin\theta = q(2l)E\sin\theta$. Substituting values, $4 = q \times 0.02 \times 2 \times 10^5 \times \sin 30^\circ$. Solving for charge gives $q = 2 \times 10^{-3} \text{ C} = 2 \text{ mC}$.
Question 32:
easyThree point charges $+q$, $-2q$ and $+q$ are placed at points $(x=0, y=a, z=0)$, $(x=0, y=0, z=0)$ and $(x=a, y=0, z=0)$ respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are: (2007)
The system can be considered as two dipoles, each of moment $p = qa$, along the x and y axes. The resultant dipole moment is $p_{net} = \sqrt{p^2 + p^2} = \sqrt{2}qa$. Its direction is along the line joining $(0,0,0)$ to $(a,a,0)$.
Question 33:
easyA dipole of dipole moment $\vec{p}$ is placed in uniform electric field $\vec{E}$ then torque acting on it is given by: (2001)
The torque acting on an electric dipole in a uniform electric field is the cross product of the dipole moment and the electric field. Thus, $\vec{\tau} = \vec{p} \times \vec{E}$.
Question 34:
easyElectric field at the equator of a dipole is $E$. If strength and distance is now doubled then the electric field will be: (1998)
The electric field at the equator of a dipole is $E \propto \frac{p}{r^3}$. When the dipole strength is doubled ($p' = 2p$) and the distance is doubled ($r' = 2r$), the new field is $E' \propto \frac{2p}{(2r)^3} = \frac{1}{4} \frac{p}{r^3} = E/4$.
Question 35:
easyA charge $q$ is placed in an uniform electric field $E$. If it is released, then the K.E. of the charge after travelling distance $y$ will be: (1998)
The force experienced by the charge is $F = qE$. The work done by this force in moving the charge through a distance $y$ is $W = F \times y = qEy$. According to the work-energy theorem, this work done is equal to the kinetic energy acquired, so K.E. = $qEy$.
Question 36:
easyA charge $q$ is located at the centre of a cube. The electric flux through any face is: (2003)
Total flux through the cube is $\frac{q}{\varepsilon_0}$ by Gauss's law. Since a cube has 6 identical faces, the flux through any one face is one-sixth of the total flux. Therefore, the flux through any face is $\frac{q}{6\varepsilon_0} = \frac{4\pi q}{6(4\pi\varepsilon_0)}$.
Question 37:
easyA charge $Q \mu\text{C}$ is placed at the centre of a cube, the flux coming out from any surface will be: (2001)
The enclosed charge is $Q \mu\text{C} = Q \times 10^{-6} \text{ C}$. By Gauss's Law, total flux through the cube is $\frac{Q \times 10^{-6}}{\varepsilon_0}$. Since a cube has 6 symmetric faces, flux through any one surface is $\frac{Q \times 10^{-6}}{6\varepsilon_0}$.
Question 38:
easyA charge $Q$ is situated at the corner of a cube, the electric flux passed through all the six faces of the cube is: (2000)
A charge placed at the corner of a cube is shared equally among 8 identical adjacent cubes. Therefore, the effective charge enclosed by the single cube is $\frac{Q}{8}$. By Gauss's law, the total electric flux passing through the cube is $\frac{Q}{8\varepsilon_0}$.
Question 39:
easyA point charge $+q$ is placed at the centre of a cube of side $l$. The electric flux emerging from the cube is (1996)
According to Gauss's Law, the total electric flux through a closed surface depends only on the net enclosed charge divided by permittivity, and is independent of dimensions. Therefore, the total flux emerging from the cube is $\frac{q}{\varepsilon_0}$.