Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A short electric dipole has a dipole moment of $16 \times 10^{-9}\, Cm .$ The electric potential due to the dipole at a point at a distance of $0.6\, m$ from the centre of the dipole, situated on a line making an angle of $60^{\circ}$ with the dipole axis is $.........V$
$\left(\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} Nm ^{2} / C ^{2}\right)$
The area of each plate of a parallel plate capacitor is $100\,c{m^2}$and the distance between the plates is $1\,mm$. It is filled with mica of dielectric $6$. The radius of the equivalent capacity of the sphere will be.......$m$
Two identical capacitors are connected in parallel across a potenial difference $V$. After they are fully charged, the positive plate of first capacitor is connected to negative plate of second and negative plate of first is connected to positive plate of other. The loss of energy will be
The dielectric strength of air at $NTP$ is $3 \times {10^6}\,V/m$ then the maximum charge that can be given to a spherical conductor of radius $3\, m$ is
A capacitor of capacitance $10\, mF$ is connected to a battery of emf $2\,V.$ It is found that it takes $50\, ms$ for the charge on the capacitor to become $12.6 \,mC.$ Then the resistance of the circuit is :.......$\, k \Omega$(Take $1/e = 0.37$):-
If some charge is given to a solid metallic sphere, the field inside remains zero and by Gauss's law all the charge resides on the surface. Now, suppose that Coulomb's force between two charges varies as $1 / r^{3}$. Then, for a charged solid metallic sphere
Two identical capacitors $1$ and $2$ are connected in series. The capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0\ volts$ . The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charge stored in the capacitors before removing the slab and $Q'_1$ , and $Q'_2$ be the values after removing the slab. Then