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.
Assertion: No work is done in moving a test charge from one point to another over an equipotential surface.
Reason: Electric field is always normal to the equipotential surface at every point.
Since potential is constant on an equipotential surface, the work done in moving a charge is zero \(W = q \Delta V = 0\). This is because the electric field is always perpendicular to the displacement along the surface \(\vec{E} \cdot d\vec{r} = 0\).
Force between two identical spheres charged with same charge is \(F \). If 50% charge of one sphere is transferred to the other sphere then the new force will be:
Initially, \( F = k \frac{q^2}{r^2}\). When 50% of the charge of one sphere is transferred to the other, the charges become \(0.5q\) and
\(1.5q\). The new force is \(F' = k \frac{(0.5q)(1.5q)}{r^2} = 0.75 F = \frac{3}{4}F\).
Assertion: The distribution of charge given to a metallic sphere does not depend on whether it is hollow or solid.
Reason: The charge resides only at the surface of conductor.
Since charges repel, they reside entirely on the outer surface of metallic conductors. Thus, the charge distribution is identical for hollow and solid metallic spheres of the same radius.
An electric dipole has the magnitude of its charge as \(q\) and its dipole moment is \(p\). It is placed in a uniform electric field \(E\). If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively:
In a uniform electric field, the net force on a dipole is always zero. The potential energy is \(U = -pE \cos(0^\circ) = -pE\), which is the minimum potential energy.
A hollow metal sphere of radius 5 cm is charged so that the potential on its surface is 10 V. The potential at the centre of the sphere is :
Inside a hollow metal sphere, the electric field is zero. This implies the electric potential is constant throughout the interior and equal to the surface potential of 10 V.
A large sphere \( P \) of radius \( R \) is charged positively. It is momentarily connected to a small sphere \( Q \) of radius \( r \). The two spheres now have same
When two conductors are connected by a conducting path (like a wire), charge flows between them until they reach electrostatic equilibrium, which means they acquire the same electrical potential.