Gauss Law/ Electric potential
The answer according to me should be 2rhor/3epsilon along the d vector
11 Replies
@Gyro Gearloose
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+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.The answer I got isn't even in the options. The correct answer is option (3)
It is option 3.
You can split the field into each sphere's contribution and add.
Each sphere's contribution is nothing but the field due the all the charge enclosed in the sphere of radius (d/2), since any spherical shell elements do not impact the field.
We take d/2 because, by symmetry, field at the centre of the region is identical to that everywhere in the region (no charge density, so boundary conditions satisfy the whole space)
You get that the field due to each sphere is ρd/6ε in the d direction (unit vector), so, taking two contributions, we get ρd/3ε (d unit vector)
But, d(d unit vector) = d vector.
So that gives you option 3
I don't understand why we take d/2
Wdym by 'no charge density so boundary conditions satisfied' @Opt
Ok, long story short, because of the options, we can assume that field is uniform within the region, and therefore, we can just consider field at the centre
Alright
+closed @Opt
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