A long cylindrical volume contains a uniformly distributed charge of density \rho \,Cm^-3 . The electric field inside the cylindrical volume at a distance x=\frac2 \varepsilon_0\rho m from its axis is \underline\hspace2cm Vm^-1 .

Electrostatics PYQ — JEE Main 2022 · 27 Jul · Shift 1 · Q68 with Solution

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Correct answer: 1

For a long solid cylinder with uniform volume charge density $\rho$ , the electric field at a distance $r$ from the axis inside the cylinder is

E(2\pi rL)=\frac{1}{\varepsilon_0}(\rho \pi r^2 L)

where:

E = \frac{\rho \pi r^2 L}{\varepsilon_0 \cdot 2\pi rL} = \frac{\rho r}{2\varepsilon_0}

So, inside the cylinder,

E = \frac{\rho r}{2\varepsilon_0}

Given

r=x=\frac{2\varepsilon_0}{\rho}

Therefore,

E=\frac{\rho}{2\varepsilon_0}\left(\frac{2\varepsilon_0}{\rho}\right)=1

E = 1\ \mathrm{V\,m^{-1}}

So the required integer is 1.