A uniformly charged disc of radius $R$ having surface charge density $\sigma$ is placed in the ${xy}$ plane with its center at the origin. Find the electric field intensity along the $z$-axis at a distance $Z$ from origin :-
A uniformly charged disc of radius $R$ having surface charge density $\sigma$ is placed in the ${xy}$ plane with its center at the origin. Find the electric field intensity along the $z$-axis at a distance $Z$ from origin :-
- [JEE MAIN 2021]
- A
${E}=\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{{Z}}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}\right)$
- B
${E}=\frac{\sigma}{2 \varepsilon_{0}}\left(1+\frac{{Z}}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}\right)$
- C
${E}=\frac{2 \varepsilon_{0}}{\sigma}\left(\frac{1}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}+{Z}\right)$
- D
${E}=\frac{\sigma}{2 \varepsilon_{0}}\left(\frac{1}{\left({Z}^{2}+{R}^{2}\right)}+\frac{1}{{Z}^{2}}\right)$
Similar Questions
A charged oil drop is suspended in a uniform field of $3 \times$ $10^{4} V / m$ so that it neither falls nor rises. The charge on the drop will be $.....\times 10^{-18}\; C$
(take the mass of the charge $=9.9 \times 10^{-15} kg$ and $g=10 m / s ^{2}$ )
A charged oil drop is suspended in a uniform field of $3 \times$ $10^{4} V / m$ so that it neither falls nor rises. The charge on the drop will be $.....\times 10^{-18}\; C$
(take the mass of the charge $=9.9 \times 10^{-15} kg$ and $g=10 m / s ^{2}$ )
- [AIEEE 2004]
The intensity of electric field required to balance a proton of mass $1.7 \times {10^{ - 27}} kg$ and charge $1.6 \times {10^{ - 19}} C$ is nearly
The intensity of electric field required to balance a proton of mass $1.7 \times {10^{ - 27}} kg$ and charge $1.6 \times {10^{ - 19}} C$ is nearly
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
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]
A charged cork of mass $m$ suspended by a light string is placed in uniform electric filed of strength $E= $$(\hat i + \hat j)$ $\times$ $10^5$ $NC^{-1}$ as shown in the fig. If in equilibrium position tension in the string is $\frac{{2mg}}{{(1 + \sqrt 3 )}}$ then angle $‘\alpha ’ $ with the vertical is
A charged cork of mass $m$ suspended by a light string is placed in uniform electric filed of strength $E= $$(\hat i + \hat j)$ $\times$ $10^5$ $NC^{-1}$ as shown in the fig. If in equilibrium position tension in the string is $\frac{{2mg}}{{(1 + \sqrt 3 )}}$ then angle $‘\alpha ’ $ with the vertical is
The distance between a proton and electron both having a charge $1.6 \times {10^{ - 19}}\,coulomb$, of a hydrogen atom is ${10^{ - 10}}\,metre$. The value of intensity of electric field produced on electron due to proton will be
The distance between a proton and electron both having a charge $1.6 \times {10^{ - 19}}\,coulomb$, of a hydrogen atom is ${10^{ - 10}}\,metre$. The value of intensity of electric field produced on electron due to proton will be