According to Gauss law of electrostatics, electric flux through a closed surface depends on
Gauss's law states that the net electric flux \(\phi\) through any closed surface is equal to \(1/varepsilon_0\) times the net charge enclosed by the surface (\(\phi = q_{\text{enclosed}}/\varepsilon_0\)), independent of the shape, area, or volume of the surface.
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 \(E = \frac{kQ}{r^2} = \frac{VR}{r^2}\).
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.
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.
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 system of two charges \(q_A = 2.5 \times 10^{-7}\text{ C}\) and \(q_B = -2.5 \times 10^{-7}\text{ C}\) are located at points A: \((0, 0, -15\text{ cm})\) and B: \((0, 0, 15\text{ cm})\) respectively. The electric dipole moment of the system is
Electric dipole moment is given by \(p = q \times 2a\), where \(2a\) is the distance between the charges. Here, \(2a = 30\text{ cm} = 0.3\text{ m}\). Thus, \(p = 2.5 \times 10^{-7}\text{ C} \times 0.3\text{ m} = 7.5 \times 10^{-8}\text{ C m}\).