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Home/Practice questions/Physics
PhysicsElectric Charges and Fields

The magnitude of the electric field in the region between the two lon…

The magnitude of the electric field in the region between the two long coaxial cylinders of a charged cylindrical capacitor

Answer & explanation+

Correct answer: option 3

The correct answer is Option (3) → varies as \( 1/r \), where \( r \) is the distance from the axis

Use Gauss’s law with a coaxial cylindrical Gaussian surface of radius \( r \) and length \( l \) between the two cylinders. This surface encloses only the charge \( \lambda l \) of the inner cylinder. The field is radial, so only the curved surface, of area \( 2\pi rl \), has flux through it.

    \( \displaystyle E\times2\pi rl=\frac{\lambda l}{\varepsilon_0} \)

    \( \displaystyle \Rightarrow\; E=\frac{\lambda}{2\pi\varepsilon_0 r} \)

Option (1) is the same throughout — Incorrect. A uniform field is found in a parallel plate capacitor. Here the field lines spread out radially, so the field gets weaker farther from the axis.

Option (2) is higher near the outer cylinder than near the inner cylinder — Incorrect. The field falls as \( r \) grows, so it is strongest near the inner cylinder and weakest near the outer one.

Option (3) varies as \( 1/r \), where \( r \) is the distance from the axis — Correct. With \( \lambda \) and \( \varepsilon_0 \) fixed, \( E\propto1/r \). This is the same as the field of a long uniformly charged straight wire.

Option (4) varies as \( 1/r^{2} \), where \( r \) is the distance from the axis — Incorrect. A \( 1/r^{2} \) field is found around a point charge or a charged sphere. There the area of the Gaussian surface grows as \( r^{2} \). The curved area of a cylinder grows only as \( r \).

NCERT (2026-27):
Field Due to an Infinitely Long Straight Uniformly Charged Wire – “… The surface includes charge equal to λl. Gauss’s law then gives

    \( E\times2\pi rl=\lambda l/\varepsilon_0 \)

i.e.,

    \( \displaystyle E=\frac{\lambda}{2\pi\varepsilon_0 r} \)

Vectorially, E at any point is given by

    \( \displaystyle \mathbf{E}=\frac{\lambda}{2\pi\varepsilon_0 r}\,\hat{\mathbf{n}} \qquad (1.32) \)

where \( \hat{\mathbf{n}} \) is the radial unit vector in the plane normal to the wire passing through the point.”

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