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Three charged particles A, B, and C with charges $-4q, 2q\: \: and\: -2q$ are present on the circumference of a circle of radius d. The charged particles A, C and the centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along the x-direction is:
Option: 1 $\frac{\sqrt{3}q}{4\pi\epsilon _0d^2}$

Option: 2 $\frac{3\sqrt{3}q}{4\pi\epsilon _0d^2}$

Option: 3 $\frac{\sqrt{3}q}{\pi\epsilon _0d^2}$

Option: 4 $\frac{2\sqrt{3}q}{\pi\epsilon _0d^2}$

Let E₁ be the resultant electric field due to charge 2q and - 2q. The resultant E₁ is in the direction of -2q

$E_1=\frac{k2q}{d^2}+\frac{k2q}{d^2}=\frac{k4q}{d^2}$

E₂ due to -4q is in the direction of -4q and magnitude is given by

$E_2=\frac{k4q}{d^2}$

The net electric field in the x-direction is

$\\E_x=E_1 cos30+E_2cos30=\frac{\sqrt{3}}{2}\times2\times\frac{4q}{4\pi\epsilon_0d^2}\\E_x=\sqrt{3}\frac{q}{\pi\epsilon_0d^2}$

So the correct option is 3.

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vishal kumar

In finding the electric field using Gauss law the formula $\left | \vec{E} \right |=\frac{q_{enc}}{\epsilon _{0}\left | A \right |}$ is applicable. In the formula $\epsilon _{0}$ is permittivity of free space, A is the area of Gaussian surface and $q _{enc}$ is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?
Option: 1 For any choice of Gaussian surface.
Option: 2 Only when the Gaussian is an equipotential surface.
Option: 3 Only when Gaussian surface is an equipotential surface and $\left | \vec{E} \right |$ is constant on the surface.
Option: 4 Only when $\left | \vec{E} \right |$ = constant on the surface.

As Gauss's law is one of the fundamental laws of physics which states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.

And

The relation of gauss's law is valid when

The magnitude of the electric field is constant & the surface is equipotential

So the correct option is 3.

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vishal kumar

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The effective capacitance of parallel combination of two capacitors C₁ and C₂ is $10\mu F$ . When these capacitors are individually connected to a voltage source of $1V$ , the energy stored in the capacitor C₂ is $4$ times that of C₁. If these capacitors are connected in series, their effective capacitance will be ( $in \ \mu F$ ):
Option: 1 1.6
Option: 2 8.4
Option: 3 3.2
Option: 4 4.2

When capacitors are in parallel,

$C_{eq}=C_1+C_2$

So, $C_1+C_2=10 \mu F$

Now when they are individually connected to 1V battery, then

The energy of C₂=4 x energy of C₁

$\Rightarrow \frac{1}{2}C_2V^2=4\times \frac{1}{2}C_1V^2$

$\Rightarrow C_2=4C_1$

So, C₂=8 and C₁=2

So when both are in series, then $C_{eq}=\frac{C_1C_2}{C_1+C_2}=\frac{8}{5}=1.6 \ \mu F$

So the correct option is 1.

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vishal kumar

An electric field $\hat{E}=4\; x\hat{i}-(y^{2}+1)\hat{j}\; N/C$ passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as F_Iand F_IIrespectively. The difference between (F_I-F_II) is (in $Nm^{2}/C$ )______.
Option: 1 48
Option: 2 38
Option: 3 58
Option: 4 62

Flux via ABCD

$\phi _{1}=\int \vec{E}\cdot d\vec{A}=0$

Flux via BCEF

$\phi _{2}=\int \bar{E}\cdot d\bar{A}$

$\phi _{2}= \bar{E}\cdot \bar{A}$

$=(4x\hat{i}-(y^{2}+1)\hat{j})\cdot 4\hat{i}$

$=16x,x=3$

$\phi_{2}=48 \frac{N-m^{2}}{C}$

$\left | \phi _{1}-\phi _{2} \right |=48\frac{N-m^{2}}{C}$

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avinash.dongre

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A parallel plate capacitor has plates of area A separated by distance'd' between them. It is filled with a dielectric which has a
dielectric constant that varies as $\mathrm{k}(x)=\mathrm{K}(1+\alpha x)$ where $^{\prime} x^{\prime}$ is the distance measured from one of the plates. If $(\alpha \mathrm{d})< < 1$ , the total capacitance of the system is best given by the expression:

Option: 1 $\frac{A\epsilon _{0}K}{d}\left ( 1+\left ( \frac{\alpha d}{2} \right )^{2} \right )$

Option: 2 $\frac{AK\epsilon _{0}}{d}\left ( 1+\frac{\alpha d}{2} \right )$

Option: 3 $\frac{A\epsilon _{0}K}{d}\left ( 1+\frac{\alpha^{2} d^{2}}{2} \right )$

Option: 4 $\frac{AK\epsilon _{0}}{d}\left ( 1+\alpha d\right )$

Dielectrics -

Dielectric: A dielectric is an insulating material in which all the electrons are tightly bounded to the nuclei of the atoms and no free electrons are available for the conduction of current. They are non-conducting materials. They do not have free charged particles like conductors have. They are of two types.

1. Polar : The centre of +ve and –ve charges do not coincide. Example HCl, $H_2O$ , They have their own dipole moment

2. Non-Polar : The centers of +ve and –ve charges coincide. Example $CO_2 , C_6H_6 .$ They do not have their own dipole moment.

When a dielectric slab is exposed to an electric field, the two charges experience force in opposite directions. The molecules get elongated and develops a surface charge density $\sigma _{p}$ . This leads to development of an induced electric field Ep , which is in opposition direction of external electric field Eo . Then net electric field E is given by $E = E_{o} - E_{i} .$

This indicates that net electric field is decreased when dielectric is introduced.

The ratio $\frac{E_0}{E}=K$ is called dielectric constant of the dielectric. Hence, Electric field inside a dielectric is $E_i=\frac{E_0}{K}$ .

$E=E_{0}-E_{i} \ and \ E=\frac{E_0}{k}\\ So,\ E_{0}-E_{i}=\frac{E_{0}}{K}$

$\begin{array}{ll}{\text { or } E_{0} K-E_{i} K=E_{0}} \\ {\text{ or } E_{0} K-E_{0}=E_{i} K} \\ {\text { or } \quad E_{i}=\frac{K-1}{K} E_{0}} & {} \\ {\text { or } \frac{\sigma_{i}}{\varepsilon_{0}}=\frac{K-1}{K} \frac{\sigma}{\varepsilon_{0}}} \\ or\ {\sigma_{i}=\frac{K-1}{K} \sigma} \\ {\text { or } \quad \frac{Q}{A}=\frac{K-1}{K} \frac{Q}{A}} \\ {\text { or } \quad Q_{i}=Q\left(1-\frac{1}{K}\right)} \end{array}$$

This is irrespective of the thickness of the dielectric slab,i.e., whether it fills up the entire space between the charged plates or any part of it.

${C}'=\frac{\epsilon _{0}A}{d-t+\frac{t}{k}}$

If K filled between the plates -

${C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck$

Given $K=K_{0}+K_0\alpha x$

$V= -\int_{0}^{d} Edr= V=\int_{0}^{d}\frac{\sigma }{K\epsilon _{0}} dx$

$V= \frac{\sigma}{\varepsilon _{0}} \int_{0}^{d}\frac{1}{K+K \alpha x} dx= \frac{\sigma }{K \alpha \varepsilon _{0} }[ln(K+K \alpha d)-lnK]$

$V= \frac{\sigma }{K \alpha \varepsilon _{0} }ln (1+ \alpha d)$

For

$ln (1+ \alpha d)=\alpha d-\frac{(\alpha d)^2}{2}+\frac{(\alpha d)^3}{3}.......$

$V= \frac{\sigma }{K \alpha \varepsilon _{0} }ln (1+ \alpha d)$

$C= \frac{Q}{V}=\frac{\sigma A}{V}= \frac{\sigma A}{\frac{\sigma }{K\varepsilon _{0} \alpha }ln (1+ \alpha d)}$

here, $C_0=\frac{\varepsilon _{0}A}{d}$

$C=\frac{Kd\alpha }{ln\left ( 1+ \alpha d \right )}C_{0}=\frac{KdC_0\alpha }{\alpha d-\frac{(\alpha d)^2}{2}+\frac{(\alpha d)^3}{3}.....}=\frac{KdC_0\alpha }{(\alpha d)*(1-\frac{\alpha d}{2}+\frac{(\alpha d)^2}{3}...)}$

since $\alpha d << 1$

$\\ C=\frac{KdC_0\alpha }{(\alpha d)*(1-\frac{\alpha d}{2} )}= \frac{KdC_0\alpha }{(\alpha d)}*(1-\frac{\alpha d}{2} )^{-1}=\frac{KdC_0\alpha }{(\alpha d)}*(1+\frac{\alpha d}{2} ) \\ \\ \\ =\frac{AK\varepsilon _0}{d}(1+\frac{\alpha d}{2} )$

So, option (2) is correct.

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Ritika Jonwal

Two infinite planes each with uniform surface charge density + $\sigma$ are kept in each such a way that the angle between them is 300. The electric field in the region shown between them is given by:
Option: 1 $\frac{\sigma }{\epsilon _{0}}\left [ \left ( 1+\frac{\sqrt{3}}{2} \right )\widehat{y}+\frac{\widehat{x}}{2} \right ]$

Option: 2 $\frac{\sigma }{2\epsilon _{0}}\left [ \left ( 1+{\sqrt{3}} \right )\widehat{y}+\frac{\widehat{x}}{2} \right ]$

Option: 3 $\frac{\sigma }{2\epsilon _{0}}\left [ \left ( 1+{\sqrt{3}} \right )\widehat{y}-\frac{\widehat{x}}{2} \right ]$

Option: 4 $\frac{\sigma }{2\epsilon _{0}}\left [ \left ( 1-\frac{\sqrt{3}}{2} \right )\widehat{y}-\frac{\widehat{x}}{2} \right ]$

Electric field due to continuous charge distribution -

So, let us consider a rod of length $l$ which has uniformly positive charge per unit length lying on x-axis, dx is the length of one small section. This rod is having a total charge Q and dq is the charge on dx segment. The charge per unit length of the rod is $\lambda$ . We have to calculate the electric field at a point P which is located along the axis of the rod at a distance of 'a' from the nearest end of Rod as shown in the figure -

The field $\vec{dE}$ at P due to each segment of charge on the rod is in the negative "x" direction because every segment of the rod carry positive charge. In this every segment of the rod is producing electric field in the negative "x" direction, so the sum of electric field can be added directly and can be integrated because all electric field lies in the same direction.

Now, here - $dq = \lambda.dx$

$\begin{array}{l}{\qquad d E=k_{\mathrm{e}} \frac{d q}{x^{2}}=k_{\mathrm{e}} \frac{\lambda d x}{x^{2}}} \\ \\{\text { The total lield at } P \text { is }} \\ \\ {\qquad E=\int_{a}^{l+a} k_{e} \lambda \frac{d x}{x^{2}}} \\ \\ {\text { If } k_{e} \text { and } \lambda=Q/l \text { are constants and can be removed from the }} {\text { integral, then }} \\ \\ {\qquad \begin{aligned} E &=k_{e} \lambda \int_{a}^{l+a} \frac{d x}{x^{2}}=k_{e} \lambda\left[-\frac{1}{x}\right]_{a}^{l+a} \\ \\ \Rightarrow k_{e} \frac{Q}{l}\left(\frac{1}{a}-\frac{1}{l+a}\right)=\frac{k_{e} Q}{a(l+a)} \end{aligned}}\end{array}$

Now if we slide the rod toward the origin and the $a\rightarrow 0$ , then due to that end, the electric field is infinite.

Assuming sheet to be non conducting

$E_{net}=E((1-\frac{\sqrt{3}}{2})\hat j-\frac{1}{2}\hat i)$

So,

$E=\frac{\sigma}{2\epsilon_0}$

So,

$E_{net}=\frac{\sigma}{2\epsilon_0}((1-\frac{\sqrt{3}}{2})\hat y-\frac{1}{2}\hat x)$

Where y and x are corresponding unit vectors

So option (4) is correct.

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Ritika Jonwal

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Consider two charged metallic spheres S₁ and S₂ of radii R₁and R₂ , respectively. The electric fields E₁/E₂ = R₁/R₂ . Then the ratio V₁(on S₁)/V₂(on S₂) of the electrostatic potentials on each sphere is :
Option: 1 $\left ( \frac{R_{1}}{R_{2}} \right )^{3}$

Option: 2 $\left( \frac{R_{2}}{R_{1}} \right )$

Option: 3 $\frac{R_{1}}{R_{2}}$

Option: 4 $\left ( \frac{R_{1}}{R_{2}} \right )^{2}$

Given

$\\\frac{E_1}{E_2}=\frac{R_1}{R_2}\\V=-E.R\\\frac{V_1}{V_2}=\frac{E_1R_1}{E_2R_2}=(\frac{R_1}{R_2})^2$

hence the correct option is (4)

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vishal kumar

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$ between them, as shown in the figure. The capacitance will be close to :
Option: 1 $\frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{4 d } \right )$

Option: 2 $\frac{\epsilon _{0}a^{2}}{d}\left ( 1 + \frac{\alpha a }{4 d } \right )$

Option: 3 $\frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{2 d } \right )$

Option: 4 $\frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{3 \alpha a }{2 d } \right )$

Take an elemental strip of length dx at a distance x from one end

So the capacitance of an elemental strip is $\mathrm{dc}=\frac{\varepsilon_{0} \mathrm{ad} \mathrm{x}}{\mathrm{d}+\alpha \mathrm{x}}$

So integrating this will give total capacitance= $C=\int \mathrm{dc}=\int_{0}^{a}\frac{\varepsilon_{0} \mathrm{ad} \mathrm{x}}{\mathrm{d}+\alpha \mathrm{x}}$

$\begin{array}{l}{\Rightarrow \quad c=\frac{\varepsilon_{0} a}{\alpha}[\ln (d+\alpha x)]_{0}^{a}} \\ {C=\frac{\varepsilon_{0} a}{\alpha} \ln \left(1+\frac{\alpha a}{d}\right) \approx \frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{\alpha a}{2 d}\right)}\end{array}$ .....(Using $\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots$ )

Hence the correct option is (3).

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vishal kumar

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A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x traveled by it is correctly given by (graphs are schematic and not drawn to scale )
Option: 1

Option: 2

Option: 3
Option: 4

$u=0,a=\frac{qE}{m}, distance=x\\ \Rightarrow using \ \ v^2=u^2+2ax\\ \Rightarrow v^2=\frac{2qE}{m}*x$

So graph would be option 3.

Hence the correct option is (3).

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vishal kumar

Consider a sphere of radius R which carries a uniform charge density $\rho$ . If a sphere of radius $\frac{R}{2}$ is carved out of it, as shown, the ratio $\frac{\left |\hat{E_A}\right |}{\left|\hat{E_B}\right |}$ of magnitude of electric field $\hat{E_A}$ and $\hat{E_B}$ , respectively, at points A and B due to the remeaning portion is:
Option: 1 $\frac{21}{34}$
Option: 2 $\frac{18}{54}$
Option: 3 $\frac{17}{54}$
Option: 4 $\frac{18}{34}$

Condider a sphere of density $-\rho$ and radius R/2 is removed from a sphere of radius R

then, Electric field at A is $E_A = E_\rho + E_{-\rho} = 0 -\frac{\rho(\frac{R}{2})}{3\epsilon_0} = \frac{\rho R}{6\epsilon_0}$

electric field at B is $E_B = E_\rho + E_{-\rho} = \frac{\rho(R)}{3\epsilon_0} - \frac{k\rho\frac{4}{3}\pi\left(\frac{R}{2} \right )^3}{\left(\frac{3R}{2} \right )^2} = \frac{17\rho R}{54\epsilon_0}$

$\left |\frac{E_A}{E_B} \right| = \frac{9}{17} \Rightarrow \frac{18}{34}$

Hence the correct option is (4).

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avinash.dongre

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Posted by

vishal kumar

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

Posted by

vishal kumar

An electric field passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as FI and FII respectively. The difference between (FI-FII) is (in )______. Option: 1 48 Option: 2 38 Option: 3 58 Option: 4 62

Posted by

avinash.dongre

JEE Main high-scoring chapters and topics

Posted by

Ritika Jonwal

Two infinite planes each with uniform surface charge density + are kept in each such a way that the angle between them is 300. The electric field in the region shown between them is given by: Option: 1 Option: 2 Option: 3 Option: 4

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