At the centre of a uniformaly charged ring :
At the centre of a uniformly charge ring electric field becomes zero because being a vector quantity it will cancel out each.
Electric potential is a scaler quantity so will be present at the centre of ring.
E = 0, V = maximum
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.
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 \(\vec{E} \cdot d\vec{r} = 0\).
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.
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.
Two charged spherical conductors of radius \(R_1\) and \(R_2\) are connected by a wire. Then the ratio of final surface charge densities of the spheres \(\sigma_1 / \sigma_2\) is
When connected, their electric potentials become equal, i.e., \(V_1 = V_2\). Since \(V = \frac{\sigma R}{\varepsilon_0}\), we get \(\sigma_1 R_1 = \sigma_2 R_2\), which gives \(\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}\).
Twenty seven drops of same size are charged at \(220\text{ V}\) each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
By volume conservation, \(R = N^{1/3} r = 27^{1/3} r = 3r\). The total charge of the combined drop is \(Q = 27q\). The potential of the bigger drop is \(V' = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9 V = 9 \times 220 = 1980\text{ V}\).
Four electric charges of \(8\ mu\text{C}\), \(5\ mu\text{C}\), \(-3\ mu\text{C}\), \(-10\ mu\text{C}\) are placed at the corners of a square of side \(\sqrt{2}\text{ m}\). The potential at the centre of the square is
The total electric potential at the centre is \(V = \sum \frac{kq_i}{r}\). Since the distance \(r\) from each corner to the centre is equal and the sum of charges \(\sum q_i = 8 + 5 - 3 - 10 = 0\), the net potential is zero.
If some positive charge is given to a solid conductor, then its potential is
The electric field inside a conductor is zero, so no work is done in moving a charge inside it. Consequently, the potential is constant and same throughout the conductor.