Asolid conducting sphere has net positive charge and radius r = 0.800 m . at a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 27.0 v . assume that v = 0 at an infinite distance from the sphere.

here we have to calculate the net positive charge present on he surface  of the conducting  sphere.

as the sphere is a conducting one the,the charge will be stored on its surface.

though the charge is present at the surface,still it will behave just like the total charge is concentrated at the centre of the sphere.

the electric field at outside of the sphere is E=

as E=

so V=

here V=27 volt,radius of sphere is 0.800 m and the point at which the potential is considered is at 1.20 m from th centre of the sphere.hence the point is at the outside of the sphere .

hence we have 27=

                            q=27× coulomb

                                =3.6× coulomb [ans]

explanation:

an enlarged virtual image is the answer

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14