WebA disc of radius R is uniformly charged with total charge Q. Find the direction and magnitude of the electric field at the point P lying a distance x from the centre of the disc along the axis of symmetry of the disc. Given- k = 1 4 π ϵ 0 σ = Q π R 2 WebJan 13, 2024 · Find the electric field of a circular thin disk of radius \(R\) and uniform charge density at a distance \(z\) above the center of the disk (Figure \(\PageIndex{4}\)) Figure \(\PageIndex{4}\): A uniformly charged disk. As in the line charge example, the field above the center of this disk can be calculated by taking advantage of the symmetry of ...

Electric field due to a charged disc #4 - YouTube

Web1 Answers. Divide the disc into concentric rings with centre of the disc taken as common centre for all rings. We know, electric field at the centre of a ring is 0. Hence electric field due to all the possible concentric rings at the centre would be 0. There would still be a charge left due to a small point charge at the centre of the disc ... WebNov 29, 2014 · July 13, 2024 by Mini Physics Find the electric field caused by a disk of radius R with a uniform positive surface charge density σ σ and total charge Q, at a point … hendrickson towing long island

electrostatics - Electric field at the centre of a charged …

WebApr 5, 2024 · Electric field – It is the space around a charge where the electrostatic force of attraction or repulsion can be felt by other charges and charged bodies. ... {R^2} + {x^2}} }}} \right]\] for the charged disc, that the electric potential is smallest near the disc and the magnitude of the electric field is maximum near the center. ... WebApr 6, 2024 · Let E be the electric field at the required point on the axis of the disc at a distance from its centre. We have to assume the charge distribution as a collection of concentric rings of charge. Let’s consider … WebFeb 15, 2015 · Then, if we want to calculate the electric field at the centre O of the hemisphere, i.e. the centre of the biggest circle that it contains, we would want to let the origin of the coordinate system to be at O. Once we do that, the electric field calculated at O is: E = σ 4 π ϵ 0 ∫ 0 2 π ∫ 0 π / 2 ∫ 0 R − ( ρ ′, ϕ ′, θ ′) ρ ′ sin ϕ d ρ ′ d ϕ ′ d θ ′. laptop hd wallpapers 4k batman