Electric field in a certain region is given by . The SI unit of and are : [30-Jan-2023 Shift 1]

B. $\text{Nm}^2 \text{C}^{-1} ; \text{Nm}^3 \text{C}^{-1}$

A. $\text{Nm}^2 \text{C}^{-1} ; \text{Nm}^3 \text{C}^{-1}$

B. $\text{Nm}^2 \text{C}^{-1} ; \text{Nm}^3 \text{C}^{-1}$

C. $\text{Nm}^2 \text{C} ; \text{Nm}^3 \text{C}$

D. $\text{Nm}^2 \text{C} ; \text{Nm}^3 \text{C}$

Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius , with distance from the centre is represented by: [8-Apr-2023 shift 1]

Electric field due to uniformly charged insulating solid sphere

Electrostatics – JEE Main Physics Practice Questions & PYQs – Page 6

Q101JEE Main 2023NAT

Three concentric spherical metallic shells ${\text{X}}, {\text{Y}}$ and ${\text{Z}}$ of radius ${\text{a}}, {\text{b}}$ and ${\text{c}}$ respectively $[ {\text{a}} < {\text{b}} < {\text{c}}]$ have surface charge densities $\sigma ,-\sigma$ and $\sigma$ , respectively. The shells ${\text{X}}$ and ${\text{Z}}$ are at same potential. If the radii of ${\text{X}}$ & ${\text{Y}}$ are $2 {\text{cm}}$ and $3 {\text{cm}}$ , respectively. The radius of shell ${\text{Z}}$ is ________ ${\text{cm}}$ . [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{q}}_{ {\text{x}}}=\sigma 4 \pi {\text{a}}^2$ $\; {\text{q}}_{ {\text{y}}}=-\sigma 4 \pi {\text{b}}^2$ $\; {\text{q}}_{ {\text{z}}}=\sigma 4 \pi {\text{c}}^2$ Potential of $y$ $\;\;{ {\text{q}}_{ {\text{x}}}}/{4 \pi \epsilon _0 {\text{a}}}+\;{ {\text{q}}_{ {\text{y}}}}/{4 \pi \epsilon _0 {\text{b}}}+\;{ {\text{q}}_{ {\text{z}}}}/{4 \pi \epsilon _0 {\text{c}}}=\;{ {\text{q}}_{ {\text{x}}}}/{4 \pi \epsilon _0 {\text{c}}}+\;{ {\text{q}}_{ {\text{y}}}}/{4 \pi \epsilon _0 {\text{c}}}+\;{ {\text{q}}_z}/{4 \pi \epsilon _0 {\text{c}}}$ $\;\;{\sigma 4 \pi {\text{a}}^2}/{ {\text{a}}}-\;{\sigma 4 \pi {\text{b}}^2}/{ {\text{b}}}+\;{\sigma 4 \pi {\text{c}}^2}/{ {\text{c}}}=4 \pi \sigma \;{ ( {\text{a}}^{2-} {\text{b}}^{2+} {\text{c}}^2) }/{ {\text{c}}}$ $\; {\text{c}}( {\text{a}}- {\text{b}}+ {\text{c}})= {\text{a}}^{2-} {\text{b}}^{2+} {\text{c}}^2$ $\; {\text{c}}( {\text{a}}- {\text{b}})+ {\text{c}}^2=( {\text{a}}+ {\text{b}})( {\text{a}}- {\text{b}})$ $\; {\text{c}}( {\text{a}}- {\text{b}})=( {\text{a}}+ {\text{b}})( {\text{a}}- {\text{b}})$ $\; {\text{c}}= {\text{a}}+ {\text{b}}=2+3$ $\; {\text{c}}=5 {\text{cm}} \;\text{ Ans. }\;$

Q102JEE Main 2023MCQ

Which of the following correctly represents the variation of electric potential (V) of a charged spherical conductor of radius (R) with radial distance (r) from the centre? [31-Jan-2023 Shift 1]

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Q103JEE Main 2023MCQ

In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor [10-Apr-2023 shift 2]

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📖 Explanation

Electrons moves in curved path because there velocity ${ {\text{u}}}^{\to }$ may make any angle $\theta$ with acceleration $\vec{a}$ between time interval of two successive collisions.Also electron moves from lower potential to higher potential.

Q104JEE Main 2023MCQ

Electric field in a certain region is given by $\vec{E}= (\frac{A}{x^2} \hat{i}+\frac{B}{y^3} \hat{j})$ . The SI unit of $A$ and $B$ are : [30-Jan-2023 Shift 1]

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📖 Explanation

$\;{ \vec{\text{E}}=\;{ {\text{A}}}/{ {\text{x}}^2} \;{ {\text{i}}}^{\^}+\;{ {\text{B}}}/{ {\text{y}}^3} \;{ {\text{j}}}^{\^}$ $\;{[\;{ {\text{A}}}/{ {\text{x}}^2}]\$ = {\text{NC}}^{-1} \Rightarrow [ {\text{A}}]= {\text{Nm}}^2 {\text{C}}^{-1}};{[;{ {\text{B}}}/{ {\text{y}}^3}]$ = {\text{NC}}^{-1} \Rightarrow [ {\text{B}}]= {\text{Nm}}^3 {\text{C}}^{-1}}$

Q105JEE Main 2023NAT

An electron revolves around an infinite cylindrical wire having uniform linear charge density $2 \times 10^{-8} {\text{C}} {\text{m}}^{-1}$ in circular path under the influence of attractive electrostatic field as shown in the figure. The velocity of electron with which it is revolving is $\times 10^6 {\text{m}} {\ s}^{-1}$ . Given mass of electron $=9 \times 10^{-31} {\text{kg}}$ [10-Apr-2023 shift 2]

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📖 Explanation

In uniform circular motion $\;F_c=m a_c$ $\;\;\text{ (q) }\;(E)=\;{ {\ mv}^2}/{ {\text{r}}}$ (e) $(\;{2 {\text{k}} \lambda }/{ {\text{r}}}) =\;{ {\ mv}^2}/{ {\text{r}}}$ $\; {\text{v}}^2=\;{( {\text{e}})(2 {\text{k}} \lambda )}/{ {\text{m}}}=\;\frac{ (1.6 \times 10^{-19}) \times 2 \times (9 \times 10^9) \times (2 \times 10^{-8}) }{9 \times 10^{-31}}$ $\;=1.6 \times 4 \times 10^{13}$ $\; {\text{V}}^2=16 \times 4 \times 10^{12} \Rightarrow {\text{v}}=8 \times 10^6 {\text{m}} / {\ s}$ Ans. 8

Q106JEE Main 2023MCQ

Two isolated metallic solid spheres of radii ${\text{R}}$ and $2 {\text{R}}$ are charged such that both have same charge density $\sigma$ . The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma ^{'}$ . The ratio $\;\frac{\sigma ^{'}}{\sigma }$ is: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{Q_1^{'}}{4 \pi \epsilon _0 R}=\;\frac{Q_2^{'}}{4 \pi \epsilon _0(2 R)}$ $\therefore \;\; Q_2^{'}=2 Q_1^{'}$ $\;Q_1^{'}+Q_2^{'}=Q_1+Q_2$ $\therefore \;\;\frac{Q_2^{'}}{2}+Q_2^{'}=20 \pi R^2 \sigma$ $\;\;\frac{3}{2} Q_2^{'}=20 \pi R^2 \sigma$ $\therefore \;\;\frac{Q_2^{'}}{4 \pi (2 R)^2}=\;\frac{2}{3} \cdot \;\frac{20 \pi R^2 \sigma }{16 \pi R^2}$ $\therefore \;\;\frac{\sigma ^{'}}{\sigma }=\;\frac{5}{6}$

Q107JEE Main 2023NAT

As shown in figure, a cuboid lies in a region with electric field $\text{E} = 2x^2 \hat{i} - 4y \hat{j} + 6 \hat{k} \text{ N/C}$ . The magnitude of charge within the cuboid is $n \epsilon_0 \text{ C}$ . The value of $n$ is ________ (if dimension of cuboid is $1 \times 2 \times 3 \text{ m}^3$ ) [30-Jan-2023 Shift 2]

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📖 Explanation

${ \vec{\text{E}}=2 {\text{x}}^2 \;{ {\text{i}}}^{\^}-4 {\text{y}} \;{ {\text{j}}}^{\^}+6 \;{ {\text{k}}}^{\^}$ $\;\varphi _{\;\text{net }\;}=-8 \times 3+2 \times 6=-12$ $\;-12=\;{ {\text{q}}}/{E_0}$ $\;| {\text{q}}|=12 E_0$

Q108JEE Main 2023MCQ

The electric field due to a short electric dipole at a large distance (r) from center of dipole on the equatorial plane varies with distance as : [15-Apr-2023 shift 1]

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📖 Explanation

$E=\;\frac{K P}{r^3} \propto \;\frac{1}{r^3}$

Q109JEE Main 2023MCQ

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : If an electric dipole of dipole moment $30 \times 10^{-5} {\text{C}} {\text{m}}$ is enclosed by a closed surface, the net flux coming out of the surface will be zero.Reason R : Electric dipole consists of two equal and opposite charges.In the light of above, statements, choose the correct answer from the options given below. [12-Apr-2023 shift 1]

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Q110JEE Main 2023NAT

Expression for an electric field is given by $\vec{E}=4000 x^2 \hat{i} \, \text{V/m}$ . The electric flux through the cube of side $20 \, \text{cm}$ when placed in electric field (as shown in the figure) is ________ $\text{V} \, \text{cm}$ . [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\text{ Flux }\;\;={ \vec{\text{E}} \cdot { \vec{\text{A}}$ $\;=4000(0 \cdot 2)^2 \;{ {\text{V}}}/{ {\text{m}}} \cdot (0 \cdot 2)^2 {\text{m}}^2$ $\;=4000 \times 16 \times 10^{-4} {\text{Vm}}$ $\;=640 {\text{Vcm}}$ Ans. 640

Q111JEE Main 2023NAT

A cubical volume is bounded by the surfaces $x=0, x=a, y=0, y=a, z=0, z=a$ . The electric field in the region is given by $\vec{E} = E_0 \hat{i}$ . Where $E_0 = 4 \times 10^4 \text{NC}^{-1} \text{m}^{-1}$ . If $a = 2 \text{cm}$ , the charge contained in the cubical volume is $Q \times 10^{-14} \text{C}$ . The value of $Q$ is ______.Take $E_0 = 9 \times 10^{-12} \text{C}^2 / \text{Nm}^2$ ) [1-Feb-2023 Shift 2]

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📖 Explanation

$\;{ \vec{\text{E}}= {\text{E}}_0 \times \;{ {\text{i}}}^{\^}$ $\;\varphi _{\;\text{net }\;}=\varphi _{ {\text{ABCD}}}= {\text{E}}_0 {\text{a}} \cdot {\text{a}}^2$ $\;\;{ {\text{q}}_{\;\text{en }\;}}/{E_0}= {\text{E}}_0 {\text{a}}^3$ $\; {\text{q}}_{ {\text{en}}}= {\text{E}}_0 E_0 {\text{a}}^3$ $\;=4 \times 10^4 \times 9 \times 10^{-12} \times 8 \times 10^{-6}$ $\;=288 \times 10^{-14} {\text{C}}$ $\; {\text{Q}}=288$ Ans. 288

Q112JEE Main 2023NAT

As shown in the figure, a configuration of two equal point charges $( {\text{q}}_0=+2 \mu {\text{C}})$ is placed on an inclined plane. Mass of each point charge is $20 {\text{g}}$ . Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height ${\text{h}}= {\text{x}} \times 10^{-3} {\text{m}}$ . The value of ${\text{x}}$ is ________. (Take $\;\frac{1}{4 \pi \epsilon _0}=9 \times 10^9 {\text{Nm}}^2 {\text{C}}^{-2}, {\text{g}}=10 {\text{ms}}^{-2}$ ) [11-Apr-2023 shift 1]

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📖 Explanation

Point charge on equilibrium is at rest. $\;\;\text{ So }\; \;\; {\text{F}}_{ {\text{e}}}= {\text{mg}} sin \;\theta$ $\;\;{ {\text{kq}}_0 \cdot {\text{q}}_0}/{ {\text{r}}^2}= {\text{mg}} sin \;30^{\circ}$ $\;\;{ {\text{kq}}_0^2}/{ (\;{ {\text{h}}}/{ sin \;30^{\circ}}) ^2}=\;{ {\text{mg}}}/{2}$ $\;\;{9 \times 10^9 \times (2 \times 10^{-6}) ^2}/{4 {\text{h}}^2}=\;\frac{20 \times 10^{-3} \times 10}{2}$ $\;\;{9 \times 4 \times 10^9 \times 10^{-12}}/{4 {\text{h}}^2}=10^{-1}$ $\; {\text{h}}^2=9 \times 10^{-2} \;\; \;\text{ Ans. }\;$ $\; {\text{h}}=0.3 {\text{m}}=300 \times 10^{-3} {\text{m}}$ $\; {\text{x}}=300 \;\; \;\text{ Ans. }\;$

Q113JEE Main 2023MCQ

Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region $E_{\text{I}}, E_{\text{II}}$ and $E_{\text{III}}$ are: [1-Feb-2023 Shift 1]

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📖 Explanation

Assuming RHS to be $\hat{n}$ $\;\vec{E}_{ {\text{I}}}=\;\frac{\sigma }{2 E_0}(-\hat{n})+\;\frac{\sigma }{2 E_0}(-\hat{n})=-\;\frac{\sigma }{E_0}\hat{n}$ $\;\vec{E}_{I I}=0$ $\;\vec{E}_{I I I}=\;\frac{\sigma }{2 E_0}(\hat{n})+\;\frac{\sigma }{2 E_0}(\hat{n})=\;\frac{\sigma }{E_0}(\hat{n})$

Q114JEE Main 2023MCQ

In a cuboid of dimension $2 {\text{L}} \times 2 {\text{L}} \times {\text{L}}$ , a charge ${\text{q}}$ is placed at the centre of the surface ' $S$ ' having area of $4 {\text{L}}^2$ . The flux through the opposite surface to ' $S$ ' is given by [29-Jan-2023 Shift 1]

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📖 Explanation

$\varphi =\;\frac{Q / \epsilon _0}{6}$ Flux passing through shaded face $=\;{ {\text{q}}}/{6 \epsilon _0}$

Q115JEE Main 2023MCQ

A point charge of $10 \mu {\text{C}}$ is placed at the origin. At what location on the ${\text{X}}$ -axis should a point charge of $40 \mu {\text{C}}$ be placed so that the net electric field is zero at ${\text{x}}=2 {\text{cm}}$ on the ${\text{X}}$ -axis ? [25-Jan-2023 Shift 2]

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📖 Explanation

$E_p=\;\frac{K \times 10}{2^2}-\;\frac{K \times 40}{ (x_0-2) ^2}=0$ $\;\frac{1}{2}=\;\frac{2}{x_0-2}$ $x_0-2=4$ ${\text{x}}_0=6 {\text{cm}}$

Q116JEE Main 2023MCQ

A $10 \mu {\text{C}}$ charge is divided into two parts and placed at $1 {\text{cm}}$ distance so that the repulsive force between them is maximum. The charges of the two parts are: [13-Apr-2023 shift 2]

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📖 Explanation

Divide $q=10 \mu$ cinto parts $(x)$ and $(q-x)$ ${\text{F}}=\;{( {\text{K}})( {\text{x}})( {\text{q}}- {\text{x}})}/{ {\text{r}}^2}$ For ${\text{F}}$ to be maximum $\;\;\frac{d F}{d x}=0$ $\;x=\;\frac{q}{2}=\;\frac{10 \mu C}{2}=5 \mu C$ $\;q-x=10 \mu C-5 \mu C=5 \mu C$

Q117JEE Main 2023NAT

An electric dipole of dipole moment is $6.0 \times 10^{-6} {\text{C}} {\text{m}}$ placed in a uniform electric field of $1.5 \times 10^3 {\text{NC}}^{-1}$ in such a way that dipole moment is along electric field. The work done in rotating dipole by $180^{\circ}$ in this field will be _______ ${\text{mJ}}$ . [8-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{W}}_{\;\text{ext }\;}= {\text{U}}_{ {\text{f}}}- {\text{U}}_{ {\text{i}}} \;\; \{ {\text{U}}=-{ {\text{P}}}^{\to } \cdot { {\text{E}}}^{\to }\}$ $\;=- {\text{PE}} \cos \pi -(- {\text{PE}} \cos 0)$ $\;=2 {\text{PE}}$ $\;=2 \times 6 \times 10^{-6} \times 1.5 \times 10^3$ $\;=18 {\text{mJ}}$

Q118JEE Main 2023MCQ

Electric potential at a point ' ${\text{P}}$ ' due to a point charge of $5 \times 10^{-9} {\text{C}}$ is $50 {\text{V}}$ . The distance of ' ${\text{P}}$ ' from the point charge is:(Assume, $\;\frac{1}{4 \pi \epsilon _0}=9 \times 10^{+9} {\text{Nm}}^2 {\text{C}}^{-2}$ ) [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{V}}=\;\frac{1}{4 \pi \epsilon _0} \;{ {\text{q}}}/{ {\text{r}}}$ $\;\Rightarrow {\text{r}}=\;\frac{9 \times 10^9 \times 5 \times 10^{-9}}{50}$ $\;\Rightarrow {\text{r}}=\;\frac{9}{10} \times 100 {\text{cm}}$ $\; {\text{r}}=90 {\text{cm}}$

Q119JEE Main 2023MCQ

Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius ${\text{R}}$ , with distance ${\text{r}}$ from the centre ${\text{O}}$ is represented by: [8-Apr-2023 shift 1]

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Q120JEE Main 2023NAT

A uniform electric field of $10 {\text{N}} / {\text{C}}$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy $0.5 {\text{eV}}$ . The length of each plate is $10 {\text{cm}}$ . The angle $(\theta )$ of deviation of the path of electron as it comes out of the field is ______ (in degree). [25-Jan-2023 Shift 1]

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📖 Explanation

$\;0.5 {\text{e}}=\;\frac{1}{2} {\ mv}_{ {\text{x}}}^2 \Rightarrow {\text{v}}_{ {\text{x}}}=\sqrt{\;{ {\text{e}}}/{ {\text{m}}}}$ $\;\;\text{ Along }\; x {\text{L}}= {\text{v}}_{ {\text{x}}} {\text{t}}=\sqrt{\;{ {\text{e}}}/{ {\text{m}}}} {\text{t}}$ $\;\;\text{ Along y }\; {\text{v}}_{ {\text{y}}}=\;{ {\text{eE}}}/{ {\text{m}}} {\text{t}}$ $\;{\text{dividing}} \;{ {\text{v}}_{ {\text{y}}}}/{ {\text{L}}}= {\text{E}} \sqrt{\;{ {\text{e}}}/{ {\text{m}}}}= {\text{Ev}}_{ {\text{x}}}$ $\;\Rightarrow {\text{Tan}} \theta =\;{ {\text{v}}_{ {\text{y}}}}/{ {\text{v}}_{ {\text{x}}}}= {\text{E}} \times {\text{L}}=10 \times 0.1=1$ $\;\theta =45^{\circ}$ $\;$