(B) For a solid sphere of radius $R$ with uniform volume charge density $\rho$,the electric field $E$ is given by:$1$. Inside the sphere $(r < R)$: $E

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The object must be a solid sphere with uniform volume charge density.

(B) For a solid sphere of radius $R$ with uniform volume charge density $\rho$,the electric field $E$ is given by:$1$. Inside the sphere $(r $2$. Outside the sphere $(r \geq R)$: $E = \frac{Q}{4\pi\epsilon_0 r^2}$,which implies $E \propto \frac{1}{r^2}$.The given graph shows $E$ increasing linearly with $r$ until a maximum value at the surface,and then decreasing as $1/r^2$ for $r > R$. This behavior is characteristic of a uniformly charged solid sphere.

Class 12 Physics Electric Charges and Fields Electric Field and usage of Gauss's Law

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The figure shows the graph of electric field $E(r)$ versus distance $(r)$ from the center of an object. Therefore,...

A
The object must be a charged conducting solid.
B
The object must be a solid sphere with uniform volume charge density.
C
The object must be a solid cube with uniform volume charge density.
D
The object must be a charged conducting sphere.

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Electric Charges and Fields

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Electric Field and usage of Gauss's Law

There is a solid sphere of radius $R$ having uniformly distributed charge throughout it. What is the relation between electric field $E$ and distance $r$ from the centre (where $r < R$)?

Easy

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Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density $+\sigma$ and $-2 \sigma$. The force experienced by a point charge $+q$ placed at the midpoint between the two plates will be:

MediumJEE MAIN 2025

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$A$ positive point charge is released from rest at a distance $r_0$ from a positive line charge with uniform density. The speed $(v)$ of the point charge,as a function of instantaneous distance $r$ from the line charge,is proportional to

DifficultJEE MAIN 2019

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Two concentric conducting thin spherical shells $A$ and $B$ having radii $r_A$ and $r_B$ $(r_B > r_A)$ are charged to $Q_A$ and $-Q_B$ $(|Q_B| > |Q_A|)$. The electric field along a line passing through the centre is:

DifficultAIIMS 2005

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Gauss's law can help in the easy calculation of the electric field due to:

Easy

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The figure shows the graph of electric field $E(r)$ versus distance $(r)$ from the center of an object. Therefore,...

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There is a solid sphere of radius $R$ having uniformly distributed charge throughout it. What is the relation between electric field $E$ and distance $r$ from the centre (where $r < R$)?

Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density $+\sigma$ and $-2 \sigma$. The force experienced by a point charge $+q$ placed at the midpoint between the two plates will be:

$A$ positive point charge is released from rest at a distance $r_0$ from a positive line charge with uniform density. The speed $(v)$ of the point charge,as a function of instantaneous distance $r$ from the line charge,is proportional to

Two concentric conducting thin spherical shells $A$ and $B$ having radii $r_A$ and $r_B$ $(r_B > r_A)$ are charged to $Q_A$ and $-Q_B$ $(|Q_B| > |Q_A|)$. The electric field along a line passing through the centre is:

Gauss's law can help in the easy calculation of the electric field due to:

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