For a uniformly charged thin spherical shell of radius \( R \) and total charge \( Q \), the electric potential inside the shell (\( r < R \)) is constant and equal to the potential at the surface:
\( V = \frac{kQ}{R} \)
where \( k \) is Coulomb's constant. This means the potential does *not* change as you move from the center towards the surface.
For points outside the shell (\( r > R \)), the electric potential is the same as that of a point charge located at the center of the shell:
\( V = \frac{kQ}{r} \)
The potential decreases as the distance \( r \) increases, following an inverse relationship.
Based on the above analysis:
This corresponds to a graph that is constant for \( r < R \) and then decreases hyperbolically for \( r > R \).
The correct graphical representation is Option (4).
You are given a dipole of charge \( +q \) and \( -q \) separated by a distance \( 2l \). A sphere 'A' of radius \( R \) passes through the centre of the dipole as shown below and another sphere 'B' of radius \( 2R \) passes through the charge \( +q \). Then the electric flux through the sphere A is
Two charges, \( q_1 = +3 \, \mu C \) and \( q_2 = -4 \, \mu C \), are placed 20 cm apart. Calculate the force between the charges.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: