The surface charge density of a thin charged disc of radius $R$ is $\sigma $. The value of the electric field at the centre of the disc is $\frac{\sigma }{{2\,{ \in _0}}}$. With respect to the field at the centre, the electric field along the axis at a distance $R$ from the centre of the disc

  • [JEE MAIN 2013]
  • A

    reduces by $70. 7\%$

  • B

    reduces by $29.3\%$

  • C

    reduces by $9.7\%$

  • D

    reduces by $14.6\%$

Similar Questions

As shown in the figure, a particle A of mass $2\,m$ and carrying charge $q$ is connected by a light rigid rod of length $L$ to another particle $B$ of mass $m$ and carrying charge $-q.$ The system is placed in an electric field $\vec E$ . The electric force on a charge $q$ in an electric field $\vec E$ is $\vec F = q \vec E $ . After the system settles into equilibrium, one particle is given a small push in the transverse direction so that the rod makes a small angle $\theta_0$ with the electric field. Find maximum tension in the rod.

Four equal positive charges are fixed at the vertices of a square of side $L$. $Z$-axis is perpendicular to the plane of the square. The point $z = 0$ is the point where the diagonals of the square intersect each other. The plot of electric field due to the four charges, as one moves on the $z-$ axis.

A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20000\, V/m$. If mass of the particle is $9.6 \times {10^{ - 16}}\,kg$, the charge on it and excess number of electrons on the particle are respectively $(g = 10\,m/{s^2})$

An infinite number of electric charges each equal to $5\, nC$ (magnitude) are placed along $X$-axis at $x = 1$ $cm$, $x = 2$ $cm$ , $x = 4$ $cm$ $x = 8$ $cm$ ………. and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at $x = 0$ is $\left( {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,N - {m^2}/{c^2}} \right)$

Charges $Q _{1}$ and $Q _{2}$ arc at points $A$ and $B$ of a right angle triangle $OAB$ (see figure). The resultant electric field at point $O$ is perpendicular to the hypotenuse, then $Q _{1} / Q _{2}$ is proportional to

  • [JEE MAIN 2020]