What will the surface charge density be if the radius of the disk is doubled but its total charge remains the same

the new surface charge density = Q/4πr²( initial surface charge density divided by 4)

Explanation:

charge density(surface) = Q/A = charge/area

let r be the initial radius of the disk

therefore, area A = πr²

charge density = Q/πr²

Now that the radius is doubled, let it be represented as R

∴ R = 2r

Recall, charge density = Q/A

A = πR = π(2r)² = 4πr²

the new surface charge density = Q/4πr²

the initial surface charge density divided by 4

learn it on the intertant so you can solve

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .