JEE Advanced Electrostatics (Non-Conductors) 2

Electricity and Magnetism Level 2

A cylindrical portion of radius $r$ is removed from a solid sphere of radius $R$ and uniform volume charge density $\rho$ in such a way that the axis of the hollow cylinder coincides with one of the diameters of the sphere. $(r<<<R)$ . Then the electric field at point $A$ is

Assumptions

$\hat{i}$ is a unit vector to the right

This is a part of My Picks for JEE Advanced 2

\dfrac{r\rho}{6\epsilon_0}\hat{i}

\dfrac{r\rho}{3\epsilon_0}\hat{i}

-\dfrac{r\rho}{3\epsilon_0}\hat{i}

-\dfrac{r\rho}{6\epsilon_0}\hat{i}

1 solution

Karan Shekhawat
Feb 25, 2015

Simple Super Position Principle !

Well Pranjal I suggest You To mention X-Y direction's .

Thanks. Added. Can you elaborate your solution?

Pranjal Jain - 6 years, 3 months ago

@Pranjal Jain can you also post a solution to this problem

Cody Martin - 6 years, 3 months ago

kindly post full solutions and not such phrases!! @karan shekhawat and @Pranjal Jain

A Former Brilliant Member - 6 years ago

Plz any soln.

Aman Bhandare - 4 years, 1 month ago

the electric field of a spherical distribuction of charge "inside it" and of a cilindrical one depend linearly on the distance from their axis, so you can use the superposition principle and consider a sphere and a cilinder with negative density.

Gabriele Manganelli - 3 years, 9 months ago

however I think the sign should be negative

Gabriele Manganelli - 3 years, 9 months ago

you must convert the volume charge density of the rod into its linear charge density

Siddhartha Krishna - 2 years, 2 months ago

JEE Advanced Electrostatics (Non-Conductors) 2

This is a part of My Picks for JEE Advanced 2

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