According to Gauss law of electrostatics, electric flux through a closed surface depends on
By Gauss's law, \(\Phi_E = \frac{q_{\text{enclosed}}}{ε_0}\). The flux depends only on the net charge enclosed inside the surface, and is independent of the size and shape of the closed surface.
A charge \(Q\) \(mu\text{C}\) is placed at the centre of a cube. The flux coming out from any one of its faces will be (in SI unit)
By Gauss's law, the total flux through the cube is \(\Phi = \frac{q_{\text{enclosed}}}{ε_0}\). Since the charge is at the center, the flux through one of the six faces is \(\Phi_1 = \frac{\Phi}{6} = \frac{Q \times 10^{-6}}{6ε_0}\).
If a conducting sphere of radius R is charged. Then the electric field at a distance \(r\) (\(r > R\)) from the centre of the sphere would be, (V = potential on the surface of the sphere)
The electric potential at the surface of a conducting sphere is \(V = \frac{kQ}{R} \implies kQ = VR\). The electric field at any point outside the sphere (\(r > R\)) is \(E = \frac{kQ}{r^2} = \frac{VR}{r^2}\).
Two particle of equal mass m and charge q are placed at a distance of 16 cm. They do not experience any force. The value of \( \frac{q}{m} \) is:
For no net force, electrostatic force must equal gravitational force: \( \frac{q^2}{4 \pi \varepsilon_0 r^2} = \frac{G m^2}{r^2} \). Rearranging this formula yields \( \frac{q^2}{m^2} = 4 \pi \varepsilon_0 G \Rightarrow \frac{q}{m} = \sqrt{4 \pi \varepsilon_0 G} \).
Two small spheres each carrying a charge q are placed r meter apart. If one of the sphere is taken around the other one in a circular path of radius r, the work done will be equal to:
The electrostatic force is conservative and directed radially. When moving in a circular path, the displacement is always perpendicular to the force, resulting in zero work done.
An \( \alpha \)-particle of mass \( 6.4 \times 10^{-27}\text{ kg} \) and charge \( 3.2 \times 10^{-19}\text{ C} \) is situated in a uniform electric field of \( 1.6 \times 10^5\text{ Vm}^{-1} \). The velocity of the particle at the end of \( 2 \times 10^{-2}\text{ m} \) path when it starts from rest is:
Using the work-energy theorem, \( qEd = \frac{1}{2}mv^2 \). Rearranging gives \( v = \sqrt{\frac{2qEd}{m}} = \sqrt{\frac{2 \times 3.2 \times 10^{-19} \times 1.6 \times 10^5 \times 2 \times 10^{-2}}{6.4 \times 10^{-27}}} = 4\sqrt{2} \times 10^5\text{ m/s} \).
The total electric flux through a cube when a charge 8q is placed at one corner of the cube is :
A charge placed at the corner of a cube is shared equally by 8 adjacent cubes. The flux through the single cube is therefore \( \Phi = \frac{Q_{total}}{8 \varepsilon_0} = \frac{8q}{8 \varepsilon_0} = \frac{q}{\varepsilon_0} \).
An electric dipole consisting of two opposite charges of \( 2 \times 10^{-6}\text{ C} \) each separated by a distance of 3 cm is placed in an electric field of \( 2 \times 10^5\text{ N/C} \). The maximum torque on the dipole will be:
Maximum torque on a dipole in a uniform electric field is \( \tau_{max} = pE = (q \cdot 2l)E \). Substituting the given values: \( \tau_{max} = (2 \times 10^{-6}\text{ C} \times 0.03\text{ m}) \times (2 \times 10^5\text{ N/C}) = 12 \times 10^{-3}\text{ Nm} \).
An elementary particle of mass m and charge +e is projected with velocity v at a much more massive fixed particle of charge Ze, where Z > 0. What is the closest possible approach of the incident particle?
At the distance of closest approach \( r_0 \), the kinetic energy is completely converted into electrostatic potential energy: \( \frac{1}{2} m v^2 = \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r_0} \). Rearranging gives \( r_0 = \frac{Ze^2}{2\pi\varepsilon_0 m v^2} \).
Assertion: Sharper is the curvature of spot on a charged body lesser will be the surface charge density at that point.
Reason: Electric field is non-zero inside a charged conductor.
Surface charge density is directly proportional to curvature \(\sigma \propto 1/R\). Inside a conductor, the electric field is zero. Hence, both statements are false.