NEET Physics MCQ: NCERT Electric Field

🎯Difficulty :Hard

🔄Ncert Concepts :Electric Field, Gauss's Law, Spherical Symmetry

Related Questions

An electron enters between two horizontal plates separated by $2\;mm$ and having a potential difference of $1000\;V.$ The force on electron is

$8 imes {10^{ - 12}}N$

$8 imes {10^{ - 14}}N$

$8 imes {10^9}N$

$8 imes {10^{14}}N$

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🔄Ncert Concepts :electric field, force on a charge, potential difference

The electric strength of air is 2 $imes {10^7}$ NC. The maximum charge that a metallic sphere of diameter 6 mm can hold is

0.5 nC

1.0 nC

2.00 nC

4.0 nC

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🔄Ncert Concepts :Electric field, Electric potential, Capacitance of a sphere

Which of the following statement is correct?

Electric field is zero on the surface of current carrying wire.

Electric field is non-zero on the axis of hollow current carrying wire

Surface integral of magnetic field for any closed surface is equal to $\;{\mu _0}$ times of total algebraic sum of current which are crossing through the closed surface

None of the above

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🎓 Ncert Level:Analyzing

🔄Ncert Concepts :Electric field, Magnetic field, Ampere's Law, Current carrying wire, Surface integral

Let ${E_a}$ be the electric field due to a dipole in its axial plane distant $'l'$ and let ${E_q}$ be the field in the equatorial plane distant $'l'$ then the relation between ${E_a}$ and ${E_q}$ will be

${E_a} = 4{E_q}$

${E_q} = 2{E_a}$

${E_a} = 2{E_q}$

${E_q} = 3{E_a}$

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🔄Ncert Concepts :Electric field due to a dipole, Axial field, Equatorial field

Electric field on the axis of a small electric dipole at a distance $'r'$ is ${E_1}$ and ${E_2}$ at a distance of $2r$ on a line of perpendicular bisector. Then

${\overrightarrow E _2} = \frac{{ - {{\overrightarrow E }_1}}}{8}$

${\overrightarrow E _2} = \frac{{ - {{\overrightarrow E }_1}}}{16}$

${\overrightarrow E _2} = \frac{{ - {{\overrightarrow E }_1}}}{4}$

${\overrightarrow E _2} = \frac{{{{\overrightarrow E }_1}}}{8}$

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🔄Ncert Concepts :Electric Dipole, Electric Field, Axial Field, Equatorial Field

Infinite charges of magnitude q each are lying at $x = 1,2,4,8..$ metre on X-axis. The value of intensity of electric field at point $x = 0$ due to these charges will be

$12 imes {10^9}{\rm{\;}}q{\rm{\;N}}{{\rm{C}}^{ - 1}}$

zero

$6 imes {10^9}{\rm{\;}}q{\rm{\;N}}{{\rm{C}}^{ - 1}}$

$4 imes {10^9}{\rm{\;}}q{\rm{\;N}}{{\rm{C}}^{ - 1}}$

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🔄Ncert Concepts :Electric Field, Superposition Principle, Geometric Series

When a dielectric material is introduced between the plates of a charged condenser then electric field between the plates

Decreases

Increases

Remain constant

First (2) and (1)

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🎓 Ncert Level:Understanding

🔄Ncert Concepts :Capacitance, Dielectric Constant, Electric Field

A solid metallic sphere has a charge $+ 3{\cal Q}.$ Concentric with this sphere is a conducting spherical shell having charge $- {\cal Q}.$ The radius of the sphere is a and that of the spherical shell is $b(b > a).$ What is the electric field at a distance $R(a < R < b)$ from the centre?

$\frac{{4{\cal Q}}}{{2\pi {\varepsilon _0}{R^2}}}$

$\frac{{3{\cal Q}}}{{4\pi {\varepsilon _0}{R^2}}}$

$\frac{{3{\cal Q}}}{{2\pi {\varepsilon _0}{R^2}}}$

$\frac{{\cal Q}}{{2\pi {\varepsilon _0}R}}$

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🔄Ncert Concepts :Gauss's Law, Electric Field due to Spherical Charge Distribution

Two conducting sphere of radii r1 and r2 are charged to the same surface charge density. The ratio of electric field near their surface is

$r_{1\;\;}^2$ / $r_2^2$

$r_2^2$ / $r_{1\;\;}^2$

$r_{1\;\;}^\;$ / $r_2^\;$

$1{\rm{ }}:{\rm{ }}1$

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