How far apart are two conducting plates that have an electric field strength of 4.50×10^{3} V/m between them, if their potential difference is 15.0 kV? (answer in m)

d = 3.3 m.

Explanation:

If the electric field created by a charge distribution on one of the plates is constant, the work done by the field on a positive test charge, it will be equal to the electrostatic force  (which is constant) times the displacement produced by this force (which has the same direction as the field) , as follows:

W = F*d = q*E*d

As the potential is the work per unit charge, we can find the potential difference (assuming that the zero potential reference is at infinity) as follows:

V = E*d

Replacing V= 15.0*10³ V, and E= 4.5*10³ V/m, we can solve for d, as follows:

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

measure their densities same density, same sub