An infinite ladder of resistors

Problem 3.152 of Irodov

The figure below shows an inifite circuit formed by the repetition of the same link, consisting of resistance R_1 = 4.0 \ \Omega and R_2 = 3.0 \ \Omega. Find the resistance of this circuit between points A and B.

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Solution

Let’s denote the resistance between the points A and B by R_{AB}. Since the circuit is infinite, removing the first R_1 and R_2 resistors gives the same arrangement back again — the arrangement is self-similar. That means, the resistance between the points C and D is just R_{AB} without the left-most R_1 and R_2 resistors, and we may redraw the circuit as shown below.

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It is now straightforward to calculate the resistance,

(1)

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Electric field in a hollow region

IIT-JEE 2007 Paper 2, Question 6

(This problem is was repeated in JEE Advanced 2015 Paper 2, Question 11)

A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

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  1. zero everywhere
  2. non-zero and uniform
  3. non-uniform
  4. zero only at its center

Related Problems:
Electric field from a sphere with a cavity
Flux from a charged shell
Charge at one corner of a cube

Solution

As explained in a related problem we can think of the given charge distribution as a …

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Charge at one corner of a cube

A charge q sits at one corner of a cube of side length a as shown in the figure below. What is the electric flux through the shaded side?

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Related problems:
Electric field from a sphere with a cavity
Electric field in a hollow region
Flux from a charged shell

Solution

Since the problem is asking us to find the electric flux it is natural to guess that we need to apply Gauss’s Law. However, we need a closed surface with appropriate symmetry to be able to use Gauss’s Law (see article). So we consider the situation shown below.

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We have placed …

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Electric field from a sphere with a cavity

A sphere of radius a is filled with positive charge with uniform density \rho. Then a smaller sphere of radius a/2 is carved out, as shown in the figure below, and left empty. What are the direction and magnitude of the electric field at A? At B?

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Related problems:
Electric field in a hollow region
Charge at one corner of a cube
Flux from a charged shell

Solution

This problem can be solved by using the principle of superposition. For instance, consider a point charge +q at some point P in space. It creates an electric field everywhere. However, if you place a …

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Flux from a charged shell

JEE Advanced 2019 Paper 1, Question 8

A charged shell of radius R carries a total charge Q. Given \Phi as the flux of electric field through a closed cylindrical surface of height h, radius r
and with its center same as that of the shell. Here, the center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct?

[\epsilon_0 is the permittivity of free space]

  1. If h > 2R and r > R then \Phi = Q/\epsilon_0
  2. If h < 8R/5 and r = 3R/5 then \Phi = 0
  3. If h > 2R and r = 3R/5 then \Phi = Q/5\epsilon_0
  4. If h > 2R and r = 4R/5 then \Phi = Q/5\epsilon_0

Related problems:
Electric field from a sphere

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