Electric Field of an Infinite Sheet with a Hole: Superposition Principle

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To find the electric field above a hole in an infinite sheet of uniform positive charge, one can use the superposition principle. This involves calculating the electric field of the infinite sheet and then adding the field created by a uniformly charged disc of negative charge that fills the hole. The electric field along the axis of symmetry of the disc is essential for this calculation. The combined fields from both the sheet and the negative disc will yield the resultant electric field above the hole. Understanding these concepts is crucial for solving the problem accurately.
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Homework Statement



If you have an infinite (non-conducting) sheet of uniform positive charge, except for a perfectly round hole in the middle, and you are trying to find the electric field above the circle, then is that the same as trying to find the electric field of the sheet with the circle filled in with negative charge (uniform) at the same spot?

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The Attempt at a Solution


 
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I think you are close. Say we have an infinite sheet of charge. You can figure the electric field for that. Now add to that sheet of charge a disc of negative charge. I think you just need to add the superimposed fields. Do you know the electric field along the axis of symmetry of a disc of uniform charge? Add that to the field of the infinite sheet.

Corrections welcome %^)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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