The electric field inside a uniformly charged sphere with positive charge Q and radius R points radially outward from the center of the sphere and has magnitude E=Qr/4πϵ0R3 at a distance r from the center. By integrating the electric field, find the potential difference between the center of the sphere and its surface. Which is at higher potential: the center or the surface?

Given that

E = Qr/(4EoR3)

Then

E = - dV/dr

So , dV = - E dr

V=integral ( from R to 0)- E Dr

V=

- integral ( from R to 0) Qr/(4πEoR3) dr

V =

- (Q/(4πEoR3)) integral ( from R to 0) r dr

Then

V = - (Q/(4πEoR3)) (R2/2)

V = - (Q/(4πEoR3)) (R2/2)

V = - Q/(4πEo(2R))

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards