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)
Potential on the surface is \(V = \frac{kQ}{R}\), so \(kQ = VR\). At \(r > R\), the electric field is \(E = \frac{kQ}{r^2} = \frac{VR}{r^2}\).
1000 identical drops of mercury are charged to a potential of \(1 \text{V}\) each. They coalesce to form a single drop, the potential of this new drop will be
The potential of a coalesced drop is related to the individual potential by \(V' = n^{2/3} V\). For \(n = 1000\), \(V' = (1000)^{2/3} \times 1 = 100 \text{V}\).
A point charge is placed at origin. Assuming potential to be zero at infinity, potential difference at two point A and B is found to be 10 V i.e., \(V_A – V_B = 10 \text{V}\). Now if the reference at infinity is changed to 10 V, then \(V_A – V_B\) will be
The potential difference between two points is independent of the reference level of potential because the change in reference shifts both potentials by the same constant value.
If a soap bubble contracts, the pressure inside bubble
The excess pressure inside a soap bubble is \(\Delta P = \frac{4T}{r}\). When it contracts, its radius \(r\) decreases, leading to an increase in the excess pressure, hence the total internal pressure increases.
Two positive charges \(q_1\) and \(q_2\) having their sum \(Q\) are placed at \(d\) distance apart. For what values of the charges, the coulomb force between them will be maximum?
Coulomb force is proportional to the product \(q_1 q_2\). Since the sum \(q_1 + q_2 = Q\) is constant, the product is maximum when the charges are equal, i.e., \(q_1 = q_2 = Q/2\).
According to Gauss’s law in electrostatics, net electric flux through a closed surface depends on
According to Gauss's law, the net electric flux through a closed surface is given by \(\Phi = \frac{q_{\text{encl}}}{\varepsilon_0}\), which only depends on the total charge enclosed.
An electric dipole is placed in a uniform electric field making an angle \(30^\circ\) with electric field and it experiences torque equal to \(\tau\) in this scenario. The minimum work done in changing the orientation from \(30^\circ\) to \(60^\circ\) is equal to
Torque is \(\tau = pE \sin 30^\circ = pE/2 \implies pE = 2\tau\). Work done is \(W = -pE(\cos 60^\circ - \cos 30^\circ) = pE(\cos 30^\circ - \cos 60^\circ) = 2\tau(\frac{\sqrt{3}}{2} - \frac{1}{2}) = (\sqrt{3}-1)\tau\).
If electric potential (in volt) in a region is expressed as \(V(x, y, z) = 2xy – yz\), the electric field (in N/C) at point \((1, 0, 1)\text{ m}\) will be equal to
Using \(\vec{E} = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)\), we get \(E_x = -2y\), \(E_y = z - 2x\), and \(E_z = y\). Substituting \((1, 0, 1)\), we get \(vec{E} = -\hat{j}\).
A particle of mass \(m\) carrying charge \(q\) is initially kept at rest at the origin. A uniform electric field \(E\) along \(x\)-axis is switched on. What will be its kinetic energy when its coordinates are \((a, b)\)?
Work done by the electric field \(\vec{E} = E\hat{i}\) is given by \(W = q\vec{E} \cdot \vec{d} = qE\hat{i} \cdot (a\hat{i} + b\hat{j}) = qEa\). By work-energy theorem, \(K_f - K_i = W \implies K_f = qEa\).
Two charged spherical conductors of radii \( R_1 \) and \( R_2 \) (\( R_1 > R_2 \)) have equal surface charge densities placed at large distance from each other. If they are connected by a conducting wire, then
The potential of a sphere with surface charge density \( \sigma \) is \( V = \frac{\sigma R}{\varepsilon_0} \). Since \( R_1 > R_2 \), \( V_1 > V_2 \). Charge flows from higher potential (larger sphere) to lower potential (smaller sphere).