The charge distribution inside a sphere of radius $R=2 a / b$ is given by the volume charge density ...

The electric field $\vec{E}$ inside a sphere with a given charge density $\rho$ is determined using the formula:$ \vec{E} = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho}{r^2} dV $For a spherical charge distribution, the electric field is expressed as:$ \vec{E} = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho}{r^2} d\left(\frac{4}{3} \pi r^3\right) dr = \frac{1}{\epsilon_0 r^2} \int_0^r \rho(r) 4 \pi r^2 dr $Given the charge density $\rho = a r^2 - b r^3$, substituting this into the equation yields:$ \vec{E} = \frac{1}{\epsilon_0 r^2} \int_0^r \left(a r^2 - b r^3\right) 4 \pi r^2 dr $To calculate the enclosed charge $Q_{\text{enc}}(r)$:$ \begin{aligned} Q_{\mathrm{enc}}(r) &= \int_0^r \left(a r^2 - b r^3\right) 4 \pi r^2 dr \\ Q_{\mathrm{enc}}(r) &= \frac{4 \pi a r^5}{5} - \frac{4 \pi b r^6}{6} \end{aligned} $The electric field becomes zero when $Q_{\text{enc}}(r) = 0$:$ \begin{aligned} \frac{4 \pi a r^5}{5} - \frac{4 \pi b r^6}{6} &= 0 \\ \frac{a r^5}{5} &= \frac{b r^6}{6} \\ r &= \frac{6 a}{5 b} \end{aligned} $Thus, the distance from the center of the sphere where the electric field goes to zero is $\frac{6a}{5b}$.

Two uniformly charged concentric thin spherical shells of radii $r_1$ and $r_2$ have charges $+Q$ an...

Inside the Inner Shell $\left(r < r_1\right)$Inside a uniformly charged shell, the electrostatic potential $V$ is constant. Since the inner shell has a charge $+Q$ and the outer shell has a charge $-Q$, and given the spherical symmetry, the potential inside the inner shell $\left(r < r_1\right)$ is:$$ V\left(r < r_1\right)=\frac{k Q}{r_1}-\frac{k Q}{r_2}=\text { constant } $$Between the Two Shells ( $r_1 \leq r < r_2$ )In this region, the potential is influenced by the charge of the inner shell $(+Q)$ and the charge of the outer shell $-Q$. The potential at a distance $r$ from the center due to a spherical shell of charge $Q$ and radius $r_i$ for $r \geq r_i$ is given by:$$ V\left(r_1 \leq r < r_2\right)=\frac{k Q}{r}-\frac{k Q}{r_2} $$Here, $\frac{k Q}{r_2}$ will be constant but $\frac{k Q}{r}$ will decrease propotional to $\frac{1}{r}$ until at $r=r_2$ when $\frac{k Q}{r}=\frac{k Q}{r_2}$$$ V\left(r=r_2\right)=0 $$Outside the Outer Shell ( $r \geq r_2$ )For $r \geq r_2$, the potentials from both shells must be considered as if they were point charges located at the center. Since the shells have charges $+Q$ and $-Q$, the net charge is zero. Therefore, the potential outside the outer shell will be zero:$$ V\left(r \geq r_2\right)=\frac{k Q}{r}-\frac{k Q}{r}=0 $$

Consider a very long cylinder of radius $a$ having a uniform positive charge density $\rho$. A spher...

The infinite cylinder with a spherical cavity can be considered the superposition of an infinite cylinder $\rho$ and sphere of charge density $-\rho$. From the principle of superposition, we have the following:$$ \begin{gathered} E_{\text {net }}=E_{\text {cylinder }}-E_{\text {sphere }} \\ E_{\text {cylinder }}=\frac{2 k \lambda}{x} \end{gathered} $$Where: $k=\frac{1}{4 \pi \epsilon_0}$$\lambda=$ linear charge density$$ \lambda=\rho \times \text { Area }=\rho \times \pi a^2 $$$$ E_{\text {cylinder }}=\frac{2 k \rho \pi a^2}{x} $$$$ E_{\text {sphere }}=\frac{k q}{x^2} $$$$ \begin{gathered} \rho=\frac{q}{4 / 3 \pi a^3} \Rightarrow q=\frac{4}{3} \rho \pi a^3 \\ E_{\text {sphere }}=\frac{4 k \rho \pi a^3}{3 x^2} \end{gathered} $$Thus,$$ E_{n e t}=\frac{2 k \rho a^2}{x}-\frac{4 k \rho \pi a^3}{3 x^2}=2 k \rho \pi a^2\left(\frac{1}{x}-\frac{2 a}{3 x^2}\right) $$For maxima/minima:$$ \begin{gathered} \frac{d E}{d x}=0 \\ \frac{d E}{d x}=\frac{-1}{x^2}-\frac{-2 a \times 2}{3 x^3}=0 \\ \frac{1}{x^2}=\frac{4 a}{3 x^3} \\ x=\frac{4 a}{3} \end{gathered} $$The other answer would be $x=0$, but we know that the electric field would be 0 at the center

A conducting sphere of radius $a$ initially contains a uniform volume charge density $\rho$. What is...

The problem involves determining the electric flux through a spherical surface of radius $2a$, centered on a point on a charged sphere's surface. To solve this, consider:The larger sphere of radius $2a$ completely encloses the original charged sphere of radius $a$.The enclosed charge $q_{\text{enc}}$ can be found using the charge density $\rho$ and the volume of the initial sphere.The formula for electric flux $\Phi$ is given by Gauss's law:$ \Phi = \frac{q_{\text{enc}}}{\epsilon_0} $The expression for the charge density $\rho$ is:$ \rho = \frac{\text{charge}}{\text{volume}} = \frac{q_{\text{enc}}}{\frac{4}{3} \pi a^3} $Solving for $q_{\text{enc}}$, we get:$ q_{\text{enc}} = \rho \cdot \frac{4}{3} \pi a^3 $Substituting $q_{\text{enc}}$ into the expression for $\Phi$, we have:$ \Phi = \frac{\rho \cdot \frac{4}{3} \pi a^3}{\epsilon_0} $Thus, the electric flux $\Phi$ through the spherical surface is:$ \Phi = \frac{4}{3 \epsilon_0} \pi a^3 \rho $

Particle A with charge $Q$ and particle B with charge $2 Q$ are fixed at positions $\vec{r}_A$ and $...

The force on particle $A$ due to particle $B\left(\vec{F}_{A B}\right)$ and the force on particle B due to particle $\mathrm{A}\left(\vec{F}_{B A}\right)$ are action-reaction pairs. According to Newton's third law of motion, these forces are equal in magnitude and opposite in direction.OrYou can also use Coulomb's law and you will find the magnitude to be the same, and for directions, you can see they are both positively charged and will move in the opposite direction from each other.

Consider two point charges $+q$ and $+2 q$ fixed on the $x-y$ plane at $(-\ell / 2,0)$ and $(+\ell /...

$$ \begin{aligned} &\text { Using the Pythagorean theorem, The distance between }-q \text { and the positive charges }\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{\frac{l^2}{4}+\frac{3 l^2}{4}}=l \end{aligned} $$Let,$$ |F|=\left|\frac{q^2}{4 \pi \epsilon_0 l^2}\right| $$Then, the forces can be redrawn as the following:$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text { Simplifying: } $$Vectorially this can be written as:$$ \frac{1}{2} F \hat{\imath}-\frac{3 \sqrt{3}}{2} F \hat{\jmath}=\frac{F}{2}(\hat{\imath}-3 \sqrt{3} \hat{\jmath}) $$On substituting value of $F$ we get,$$ F_{n e t}=\frac{q^2}{8 \pi \epsilon_0 l^2}(\hat{\imath}-3 \sqrt{3} \hat{\jmath}) $$Using $F=m a, a=\frac{F_{\text {net }}}{m}$

Electrostatics - IAT-IISER Physics PYQs

Q: Consider a very long cylinder of radius $a$ having a uniform positive charge density $\rho$. A spher...

The infinite cylinder with a spherical cavity can be considered the superposition of an infinite cylinder $\rho$ and sphere of charge density $-\rho$. From the principle of superposition, we have the following:$$ \begin{gathered} E_{\text {net }}=E_{\text {cylinder }}-E_{\text {sphere }} \\ E_{\text {cylinder }}=\frac{2 k \lambda}{x} \end{gathered} $$Where: $k=\frac{1}{4 \pi \epsilon_0}$$\lambda=$ linear charge density$$ \lambda=\rho \times \text { Area }=\rho \times \pi a^2 $$$$ E_{\text {cylinder }}=\frac{2 k \rho \pi a^2}{x} $$$$ E_{\text {sphere }}=\frac{k q}{x^2} $$$$ \begin{gathered} \rho=\frac{q}{4 / 3 \pi a^3} \Rightarrow q=\frac{4}{3} \rho \pi a^3 \\ E_{\text {sphere }}=\frac{4 k \rho \pi a^3}{3 x^2} \end{gathered} $$Thus,$$ E_{n e t}=\frac{2 k \rho a^2}{x}-\frac{4 k \rho \pi a^3}{3 x^2}=2 k \rho \pi a^2\left(\frac{1}{x}-\frac{2 a}{3 x^2}\right) $$For maxima/minima:$$ \begin{gathered} \frac{d E}{d x}=0 \\ \frac{d E}{d x}=\frac{-1}{x^2}-\frac{-2 a \times 2}{3 x^3}=0 \\ \frac{1}{x^2}=\frac{4 a}{3 x^3} \\ x=\frac{4 a}{3} \end{gathered} $$The other answer would be $x=0$, but we know that the electric field would be 0 at the center

Q: A conducting sphere of radius $a$ initially contains a uniform volume charge density $\rho$. What is...

The problem involves determining the electric flux through a spherical surface of radius $2a$, centered on a point on a charged sphere's surface. To solve this, consider:The larger sphere of radius $2a$ completely encloses the original charged sphere of radius $a$.The enclosed charge $q_{\text{enc}}$ can be found using the charge density $\rho$ and the volume of the initial sphere.The formula for electric flux $\Phi$ is given by Gauss's law:$ \Phi = \frac{q_{\text{enc}}}{\epsilon_0} $The expression for the charge density $\rho$ is:$ \rho = \frac{\text{charge}}{\text{volume}} = \frac{q_{\text{enc}}}{\frac{4}{3} \pi a^3} $Solving for $q_{\text{enc}}$, we get:$ q_{\text{enc}} = \rho \cdot \frac{4}{3} \pi a^3 $Substituting $q_{\text{enc}}$ into the expression for $\Phi$, we have:$ \Phi = \frac{\rho \cdot \frac{4}{3} \pi a^3}{\epsilon_0} $Thus, the electric flux $\Phi$ through the spherical surface is:$ \Phi = \frac{4}{3 \epsilon_0} \pi a^3 \rho $

Q: Particle A with charge $Q$ and particle B with charge $2 Q$ are fixed at positions $\vec{r}_A$ and $...

The force on particle $A$ due to particle $B\left(\vec{F}_{A B}\right)$ and the force on particle B due to particle $\mathrm{A}\left(\vec{F}_{B A}\right)$ are action-reaction pairs. According to Newton's third law of motion, these forces are equal in magnitude and opposite in direction.OrYou can also use Coulomb's law and you will find the magnitude to be the same, and for directions, you can see they are both positively charged and will move in the opposite direction from each other.

Q: Consider two point charges $+q$ and $+2 q$ fixed on the $x-y$ plane at $(-\ell / 2,0)$ and $(+\ell /...

$$ \begin{aligned} &\text { Using the Pythagorean theorem, The distance between }-q \text { and the positive charges }\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{\frac{l^2}{4}+\frac{3 l^2}{4}}=l \end{aligned} $$Let,$$ |F|=\left|\frac{q^2}{4 \pi \epsilon_0 l^2}\right| $$Then, the forces can be redrawn as the following:$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text { Simplifying: } $$Vectorially this can be written as:$$ \frac{1}{2} F \hat{\imath}-\frac{3 \sqrt{3}}{2} F \hat{\jmath}=\frac{F}{2}(\hat{\imath}-3 \sqrt{3} \hat{\jmath}) $$On substituting value of $F$ we get,$$ F_{n e t}=\frac{q^2}{8 \pi \epsilon_0 l^2}(\hat{\imath}-3 \sqrt{3} \hat{\jmath}) $$Using $F=m a, a=\frac{F_{\text {net }}}{m}$

01

PYQ 2020

easy

physics ID: iat-iise

The charge distribution inside a sphere of radius is given by the volume charge density, where and are constants and is the radial distance from the center of the sphere. At what distance from the center of the sphere, will the electric field go to zero?

1

2

3

4

02

PYQ 2022

easy

physics ID: iat-iise

Two uniformly charged concentric thin spherical shells of radii and have charges and, respectively. How does the electrostatic potential vary with distance from the center of the shells?

03

PYQ 2022

hard

physics ID: iat-iise

Consider a very long cylinder of radius having a uniform positive charge density . A sphere of radius has been carved out (see figure), leaving no charge in that region. The distance radially outward to the cylinder, as measured from the center of the sphere is . At what value of will the electric field be maximum?

1

2

3

4

04

PYQ 2022

easy

physics ID: iat-iise

A conducting sphere of radius initially contains a uniform volume charge density . What is the electric flux through a spherical surface of radius centered at a point on the surface of the charged sphere (see figure) at a later time ? (is the permittivity of free space.)

1

2

3

4

05

PYQ 2023

medium

physics ID: iat-iise

Particle A with charge and particle B with charge are fixed at positions and, respectively. The force on due to is, and that on due to is . Which of the following options is correct?

1

2

3

4

06

PYQ 2024

easy

physics ID: iat-iise

Consider two point charges and fixed on the plane at and respectively. Another point charge having mass is released from rest at on the plane, as shown in the figure The permittivity of free space is . What is the acceleration of the charge at the time of release?

1

2

3

4

07

PYQ 2025

hard

physics ID: iat-iise

A particle of charge and mass with kinetic energy enters an electric field set up by two parallel plates of length as illustrated in the figure. The potential difference between the two plates is 1 V and their separation is . What is the minimum value of (in eV ) for which the particle will not hit either of the plates? [is the charge of the electron.]

1

2

3

4

Electrostatics

High-Yield Trend

Questions
7 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Electrostatics

High-Yield Trend

Questions 7 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Questions
7 MCQs