🛠️ JEE🧲 Physics

A uniformly charged disc of radius \(R\) having surface charge density \(\sigma\) is placed in the \(x y\) plane with it…

Q1

A uniformly charged disc of radius \(R\) having surface charge density \(\sigma\) is placed in the \(x y\) plane with its center at the origin. Find the electric field intensity along the z-axis at a distance \(Z\) from origin:

[JEE Main 2022, 24 June (Shift 1)]

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

\(E=\frac{\sigma}{2 \varepsilon_0}\left(\frac{1}{\left(Z^2+R^2\right)}+\frac{1}{Z^2}\right)\)

🔒
Answer & explanation — PYQ Pass

Options are free to see. Unlock the correct answer and full explanation with Pass.

Unlock · ₹149

Practice more JEE Physics PYQs

See every question on Electric Charges and Fields, or browse the full JEE question bank.

See all questions on Electric Charges and Fields →