In a particle accelerator, a current of $500 \,\mu A$ is carried by a proton beam in which each proton has a speed of $3 \times 10^7 \,m / s$. The cross-sectional area of the beam is $1.50 \,mm ^2$. The charge density in this beam (in $C / m ^3$ ) is close to

Three identical dipoles are arranged as shown below. What will be the net electric field at $M$

Two charges  $ + 3.2\, \times \,{10^{ – 19}}\,C$ and $ – 3.2\, \times \,{10^{ – 19}}\,C$ kept $2.4\,\mathop A\limits^o $ apart forms a dipole. If it is kept in uniform electric field of intensity $4\, \times \,{10^{5\,}}\,volt/m$ then what will be its potential energy in stable equilibrium

The electric potential $(V)$ as a function of distance $(x)$ [in meters] is given by $V = (5x^2 + 10 x -9)\, Volt$. The value of electric field at $x = 1\, m$ would be……$Volt/m$

Charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively which are at distance $2\,L$ apart, $C$ is the midpoint between $A$ and $B$ . The work done in moving a charge $+ Q$ along the semicircle $CRD$ is