




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 \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to: