Q121JEE Main 2023MCQ

The electric potential at the centre of two concentric half rings of radii $R_1$ and $R_2$ , having same linear charge density $\lambda$ is [24-Jan-2023 Shift 2]

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📖 Explanation

Potential at centre $\;V=\;{ (\lambda \cdot \pi {\text{R}}_2) }/{4 \pi \epsilon _0 {\text{R}}_2}+\;{ (\lambda \cdot \pi {\text{R}}_1) }/{4 \pi \epsilon _0 {\text{R}}_1}$ $\;=\;\frac{\lambda }{2 \epsilon _0}$

Q122JEE Main 2023NAT

Three point charges $q,-2 q$ and $2 q$ are placed on $x$ -axis at a distance $x=0, x=\;\frac{3}{4} R$ and $x=R$ respectively from origin as shown. If ${\text{q}}=2 \times 10^{-6} {\text{C}}$ and ${\text{R}}=2 {\text{cm}}$ , the magnitude of net force experienced by the charge $-2 {\text{q}}$ is _______ N. [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{F}}_{ {\text{BA}}}=\;{32 {\text{kq}}^2}/{9 {\text{q}}^2} \;\; {\text{F}}_{ {\text{BC}}}=\;{64 {\text{kq}}^2}/{ {\text{R}}^2}$ $\; {\text{F}}_{ {\text{B}}}= {\text{F}}_{ {\text{BC}}}- {\text{F}}_{ {\text{BA}}}=\;{544 {\text{kq}}^2}/{9 {\text{R}}^2}$ $\;=5440 {\text{N}}$

Q123JEE Main 2023MCQ

As shown in the figure, a point charge $Q$ is placed at the centre of conducting spherical shell of inner radius $a$ and outer radius $b$ . The electric field due to charge Q in three different regions I, II and III is given by : $( {\text{I}}: {\text{r}} < {\text{a}}$ , II : ${\text{a}} < {\text{r}} < {\text{b}}$ , III : $r > b$ ) [30-Jan-2023 Shift 2]

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Q124JEE Main 2023NAT

For a charged spherical ball, electrostatic potential inside the ball varies with ${\text{r}}$ as ${\text{V}}=2 {\text{ar}}^{2+} {\text{b}}$ .Here, ${\text{a}}$ and ${\text{b}}$ are constant and ${\text{r}}$ is the distance from the center. The volume charge density inside the ball is $-\lambda {\text{a}} \epsilon$ . The value of $\lambda$ is _______. $\epsilon =$ permittivity of medium. $\lambda =12$ [29-Jan-2023 Shift 2]

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📖 Explanation

${\text{E}}=-\;{ {\text{dV}}}/{ {\text{dr}}}=-4 {\text{ar}} \equiv \;{\rho {\text{r}}}/{3 \epsilon _0} \;\text{ (compare) }\;$ Result inside uniformly charged solid sphere. $\rho =-12 a \epsilon _0$

Q125JEE Main 2023MCQ

For a uniformly charged thin spherical shell, the electric potential $( {\text{V}})$ radially away from the entre $(O)$ of shell can be graphically represented as - [6-Apr-2023 shift 1]

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📖 Explanation

For $r \le R, V=\;\frac{K Q}{R}$ , i.e., Constant everywhere inside.For $r > R, V=\;\frac{K Q}{r}$ , i.e., Decreases with ${\text{r}}$ .

Q126JEE Main 2023NAT

64 identical drops each charged upto potential of $10 {\text{mV}}$ are combined to form a bigger drop. The potential of the bigger drop will be _______ ${\text{mV}}$ . [12-Apr-2023 shift 1]

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📖 Explanation

We know $V=\;\frac{k q}{r}, \;\; q^{'}=64 q$ Voulme remain const so $\;\;\frac{4}{3} \pi r^3 \times 64=\;\frac{4}{3} \pi R^3$ $\;R=4 r$ Now new potential ${\text{V}}^{'}=\;{ {\text{K}} 64 {\text{q}}}/{4 {\text{r}}}=\;{16 {\text{kq}}}/{ {\text{r}}}$ $\; {\text{V}}^{'}=16 . {\text{V}} \;\text{ and }\; {\text{V}}=\;{ {\text{Kq}}}/{ {\text{r}}}=10 {\text{mV}}$ $\; {\text{V}}^{'}=16 \times 10 \times 10^{-3}$ $\; {\text{V}}^{'}=0.16 {\text{V}}$ OR $V^{'}=160 {\text{mV}}$

Q127JEE Main 2023MCQ

A dipole comprises of two charged particles of identical magnitude $q$ and opposite in nature. The mass ' $m$ ' of the positive charged particle is half of the mass of the negative charged particle. The two charges are separated by a distance 'l'. If the dipole is placed in a uniform electric field ' $\overline{ {\text{E}}}$ '; in such a way that dipole axis makes a very small angle with the electric field, ' $\overline{ {\text{E}}}$ '. The angular frequency of the oscillations of the dipole when released is given by : [6-Apr-2023 shift 2]

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📖 Explanation

In this case, since masses of both charges are not same, therefore, we need to find center of mass (COM), about which dipole will oscillate and then we will find moment of Inertia about this axis, to find torque & hence $\omega$ . As we know, COM will divide length in the inverse ratio of the masses, therefore, COM will be at a distance of $\;{ {\text{L}}}/{3}$ from $2 {\text{m}} & \;{2 {\text{L}}}/{3}$ from ${\text{m}}$ MI about this axis $I=2 m (\;\frac{L}{3}) ^{2+} (\;\frac{2 L}{3}) ^2$ Or I $=\;{2 {\text{mL}}^2}/{ {\text{a}}}+\;{4 {\text{mL}}^2}/{ {\text{a}}}=\;{6 {\text{mL}}^2}/{ {\text{a}}}=\;{2 {\text{mL}}^2}/{3}$ Using $\omega =\;{2 {\text{mL}}^2}/{3} & {\text{p}}= {\text{qL}}$ $\omega =\sqrt{\;{ {\text{qLE}}}/{{2 {\text{L}}^2}/{3}}}=\sqrt{\;{3 {\text{qE}}}/{2 {\text{mL}}}}$ None of these given option is correct. (BONUS)

Q128JEE Main 2023MCQ

Two charges each of magnitude $0.01 {\text{C}}$ and separated by a distance of $0.4 {\text{mm}}$ constitute an electric dipole. If the dipole is placed in an uniform electric field ' ${ {\text{E}}}^{\to }$ ' of 10 dyne/C making $30^{\circ}$ angle with ${ {\text{E}}}^{\to }$ , the magnitude of torque acting on dipole is [13-Apr-2023 shift 1]

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📖 Explanation

Dipole moment, ${\text{P}}= {\text{qd}}$ $\; {\text{P}}=0.01 \times 0.4 \times 10^{-3}$ $\; {\text{P}}=4 \times 10^{-6} {\text{C}}- {\text{m}}$ Torque, $\tau \;=p E sin \;\theta$ $\tau \;=4 \times 10^{-6} \times (10 \times 10^{-5}) \times sin \;30^{\circ}$ $\tau \;=4 \times 10^{-10} {\text{N}}- {\text{m}}$

Q129JEE Main 2023MCQ

If two charges ${\text{q}}_{1}$ and ${\text{q}}_{2}$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant ${\text{K}}$ . What will be the equivalent distance between charges in air for the same electrostatic force ? [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{F}}=\;\frac{1}{ (4 \pi \epsilon _0) } \;{ {\text{q}}_1 {\text{q}}_2}/{ {\text{kd}}^2} \;\text{ (\in medium) }\;$ $\; {\text{F}}_{\;\text{Air }\;}=\;\frac{1}{4 \pi \epsilon _0} \;{ {\text{q}}_1 {\text{q}}_2}/{ {\text{d}}^{' 2}}$ $\; {\text{F}}= {\text{F}}_{ {\text{Air}}}$ $\;\;{ {\text{q}}_1 {\text{q}}_2}/{4 \pi \epsilon _0 {\text{kd}}^2}=\;{ {\text{q}}_1 {\text{q}}_2}/{4 \pi \epsilon _0 {\text{d}}^{' 2}}$ $\; {\text{d}}^{'}= {\text{d}} \sqrt{ {\text{k}}}$

Q130JEE Main 2023NAT

Two equal positive point charges are separated by a distance $2 a$ . The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $q_0$ becomes maximum is $\;{a}/{{\sqrt{x}}}$ . The value of $x$ is _______. [1-Feb-2023 Shift 1]

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📖 Explanation

$F=\;\frac{2 K q q_0 x}{ (x^{2+}a^2) ^{3 / 2}}$ For $F$ to be maximum $\;{ {\text{d}} F}/{ {\text{d}} x}=0$ $x=\;{a}/{{\sqrt{2}}}$

Q131JEE Main 2022MCQ

If the electric potential at any point $(x, y, z) m$ in space is given by $V=3 x^2$ volt. The electric field at the point $(1,0,3) {\text{m}}$ will be : [29-Jun-2022-Shift-2]

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📖 Explanation

$\vec{E}=-\;\frac{d V}{d x} \hat{i}$ $\vec{E}=-6 x \hat{i}$ So, $\vec{E}$ at $(1,0,3)$ is $\vec{E}=-6 \hat{i} {\text{V}} / {\text{m}}$

Q132JEE Main 2022MCQ

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6} {\text{C}}$ and $-8 \times 10^{-6} {\text{C}}$ respectively are placed at a distance ${\text{d}}$ apart. The electric field at the middle point ${\text{O}}$ between the charges is $6.4 \times 10^4 {\text{NC}}^{-1}$ . The distance ' ${\text{d}}$ ' between the point charges ${\text{A}}$ and $B$ is : [28-Jun-2022-Shift-2]

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📖 Explanation

Electric field at ${\text{P}}$ will be $E=\;\frac{k q}{(d / 2)^2} \times 2=\;\frac{8 k q}{d^2}$ So, $\;\frac{8 \times 9 \times 10^9 \times 8 \times 10^{-6}}{d^2}=6.4 \times 10^4$ So, $d=3 {\text{m}}$

Q133JEE Main 2022MCQ

Three identical charged balls each of charge $2 {\text{C}}$ are suspended from a common point $P$ by silk threads of $2 {\text{m}}$ each (as shown in figure). They form an equilateral triangle of side $1 {\text{m}}$ .The ratio of net force on a charged ball to the force between any two charged balls will be: [27-Jun-2022-Shift-2]

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📖 Explanation

$F_{\;\text{net }\;}$ on charge $3, F_1=\;{{\sqrt{3}} k q^2}/{1^2}$ Force between any 2 charges $F_2=\;\frac{k q^2}{1^2}$ So, $\;\frac{F_1}{F_2}={\sqrt{3}}$

Q134JEE Main 2022MCQ

Given below are two statements. Statement I : Electric potential is constant within and at the surface of each conductor. Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point. In the light of the above statements, choose the most appropriate answer from the options given below. [29-Jul-2022-Shift-1]

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📖 Explanation

(Properties of conductor)Statement - I, true as body of conductor acts as equipotential surface.Statement $-2$ True, as conductor is equipotential.Tangential component of electric field should be zero. Therefore electric field should be perpendicular to surface.

Q135JEE Main 2022MCQ

The three charges ${\text{q}} / 2, {\text{q}}$ and ${\text{q}} / 2$ are placed at the corners ${\text{A}}, {\text{B}}$ and ${\text{C}}$ of a square of side 'a' as shown in figure. The magnitude of electric field (E) at the corner $D$ of the square, is : [28-Jun-2022-Shift-1]

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📖 Explanation

$|E_0{{}/{}|} =\;\frac{k q / 2}{a^2} {\sqrt{2}}+\;{k q}/{(a {\sqrt{2}})^2}$ $=\;{k q}/{{\sqrt{2}} a^2}+\;\frac{k q}{2 a^2}$ $=\;\frac{k q}{a^2} (\;{1}/{{\sqrt{2}}}+\;\frac{1}{2}) , k=\;\frac{1}{4 \pi \epsilon _0}$

Q136JEE Main 2022MCQ

A spherically symmetric charge distribution is considered with charge density varying as Where, $r(r < R)$ is the distance from the centre 0 (as shown in figure). The electric field at point ${\text{P}}$ will be: [29-Jul-2022-Shift-1]

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📖 Explanation

By Gauss law $∮ { \vec{\text{E}}{\text{d}} { \vec{\ s}=\;{ {\text{Q}}_{ {\in }}}/{\epsilon _{ {\text{o}}}}$ ${\text{E}} .4 \pi {\text{r}}^2=\;{∫_{0}^{{} {\text{r}}} \rho _{ {\text{o}}} (\frac{3}{4}-{ {\text{r}}}/{ {\text{R}}}) 4 \pi {\text{r}}^2 {\text{dr}}}/{\epsilon _0}$ ${\text{E}} 4 \pi {\text{r}}^2=\;{\rho _{ {\text{o}}} 4 \pi }/{\epsilon _{ {\text{o}}}} (\;\frac{3}{4} \;{ {\text{r}}^3}/{3}-\;{ {\text{r}}^4}/{4 {\text{R}}})$ ${\text{E}} {\text{r}}^2=\;{\rho _{ {\text{o}}} {\text{r}}^3}/{4 \epsilon _{ {\text{o}}}} \{1-\;{ {\text{r}}}/{ {\text{R}}}\}$ ${\text{E}}=\;{\rho _{ {\text{o}}} {\text{r}}}/{4 \epsilon _{ {\text{o}}}} \{1-\;{ {\text{r}}}/{ {\text{R}}}\}$

Q137JEE Main 2022MCQ

In the figure, a very large plane sheet of positive charge is shown. $P_1$ and $P_2$ are two points at distance I and $2 \mid$ from the charge distribution. If $\sigma$ is the surface charge density, then the magnitude of electric fields ${\text{E}}_1$ and ${\text{E}}_2$ at $P_1$ and $P_2$ respectively are : [25-Jun-2022-Shift-1]

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📖 Explanation

For an infinite charged plane $E=\;\frac{\sigma }{2 \epsilon _0}$ for any value of ${\text{I}}$ $\Rightarrow E_1=E_2=\;\frac{\sigma }{2 \epsilon _0}$

Q138JEE Main 2022MCQ

Given below are two statements :Statement I : A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive.Statement II : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero.Choose the correct answer from the options given below : [28-Jun-2022-Shift-1]

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📖 Explanation

As one moves closer to a positive charge (isolated) the density of electric field line increases and so does the electric field intensity $\Rightarrow$ Statement I is trueAs opposite poles of an electric dipole would experience equal and opposite forces so net force on a dipole in a uniform electric field will be zero $\Rightarrow$ Statement II is true

Q139JEE Main 2022MCQ

Two identical positive charges $Q$ each are fixed at a distance of '2a' apart from each other. Another point charge $q_0$ with mass ' $m$ ' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge ${\text{q}}_0$ executes ${\text{SHM}}$ . The time period of oscillation of charge ${\text{q}}_0$ will be : [27-Jul-2022-Shift-1]

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📖 Explanation

$(x< < a)(\alpha$ is acceleration $)$ $F_{n e t}=-(\;\frac{k q_0 Q}{(a-x)^2}-\;\frac{k Q q_0}{(a+x)^2})$ $m \alpha =-\;\frac{k q_0 Q}{a^4} 4 a x$ $\Rightarrow \alpha =-\;\frac{4 k q_0 Q}{m a^3} x$ $\;\text{ So, }\; T=2 \pi \sqrt{\;\frac{4 \pi \epsilon _0 m a^3}{4 q_0 Q}}$ $\;\text{ or }\; T=\sqrt{\;\frac{4 \pi ^3 \epsilon _0 m a^3}{q_0 Q}}$

Q140JEE Main 2022MCQ

A long cylindrical volume contains a uniformly distributed charge of density $\rho$ . The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is: [24-Jun-2022-Shift-2]

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📖 Explanation

$\;\frac{m v^2}{r}=\;\frac{2 k \rho \times \pi R^2 q}{r}$ $\Rightarrow \;\frac{1}{2} m v^2=\;\frac{\rho R^2 q}{4 \epsilon _0}$