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enter image description here This question was asked in IIT JEE 2010. I know that the electrostatic pressure is equal to (1/2)∈(E)² and if we multiply this with the projected area then we would get the force acting on an object holding charge. In the solutions of this question the electric field is taken to be (σ/ϵ) as the electric field on the surface of any one of the hemispherical conductors is (σ/ϵ) and then the total force acting on the other hemispherical shell is found out by multiplying that electrostatic pressure with the projected area , i.e. , (π*(r)^2). But I do not understand that why the electric field is taken to be (σ/ϵ) . Though the electric field due to any conductor on it's surface is equal to (σ/ϵ), it is not constant throughout the surface of the other hemispherical shell. Therefore , the electrostatic pressure should not be constant throughout the surface of any one of the shells.

So , how can we use this formula for finding the electric force on the complete body?

σ - Charge density of the hemispherical shells; ϵ - permittivity of free space; E - Electric field

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E=σ/ϵ is derived from guass law.E is the field due to spherical shell.E is constant all over spherical shell as charge is distributed uniformly.(Imagine small points charges placed equally away.Wouldn't their field be constant at any point on sphere?)

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  • $\begingroup$ I know that E is constant all over the surface of the shell due to which it is being produced. But suppose the shell 1 is producing the electric field E and repelling the shell 2. What I am saying is that the electric field due to 1 on the surface of 2 should not be constant. But the solutions to this question have used the value of E as (σ/ϵ) , which is true only on the surface of 1. On the surface of 2 , the electric field due to 1 should have different values at each point , shouldn't it? $\endgroup$ – Pratham Yadav May 25 at 17:13
  • $\begingroup$ Think like this -Note that radius of each sphere is same.Thus net charge on both shell is same. Let's say if q1 is charge on shell 1 and is repelling charge on shell 2 (q2) wouldn't q2 do the same(as q1 =q2 the magnitude of field will be equal) $\endgroup$ – Bhavay May 25 at 17:17
  • $\begingroup$ Yes , shell 2's behaviour would be the exact same as that of shell 1. But what I am asking is that how can we put the same value of Electric field in the formula for electrostatic pressure on shell 2 , if the electric field due to 1 is different on every point on the surface of shell 2? $\endgroup$ – Pratham Yadav May 25 at 17:20
  • $\begingroup$ Shell 1 is exerting a force on shell 2 Lets assume it to be X newton.Now shell 2 by symmetry should also exert a force on all charges of shell 1 =X. Shell 1 E = 𝜎^2𝜋𝑅^2/2𝜖0 thus total force will be 𝜎^2𝜋𝑅^2/𝜖0.(Note that for shell1 field/force is due to shell 2 as shell 1 can't exist field on itself ) $\endgroup$ – Bhavay May 25 at 17:30
  • $\begingroup$ Sir , I think that you are not getting my question . I am saying that shell 1 is exerting a force on shell 2 through its electric field , and shell 2 is exerting the same force on shell 1 due to symmetry . But what I am not being able to understand is that how can the electrostatic pressure on one shell , due to the other shell , be constant throughout it's surface. I mean , the electric field due to shell 1 is constant only on its surface , not anywhere else . Thus , throughout the surface of shell 2 , the electric field due to shell 1 must be continuously changing , and vice versa. $\endgroup$ – Pratham Yadav May 25 at 17:38

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