Question

1 A capacitor is constructed from a pair of thin square plates of side-width /(i.e. Area-P) and separation 3d. Let's assume 3d<<I. A solid square conducting plate of thickness d with the same side-width I (i.e. Area - P) is inserted between the plates as shown below Charge +0 is supplied on the top plate and-0 is supplied on the bottom plate. What is the electric field E between the conductors? (Express E by Q.d l and &) a) b) Draw the electric field lines in this picture -Q Square (IxI) c) Explain the charge distribution in and on the middle conductor.

Answer 1

A capacitor is constructed from a pair of this square plates of side-width 1(i.e Area = 1^2) and separation 3d a E_1 = sigma/elementof_0 (OR) E_1 = Q/l^2 elementof_0 And elementof_3 = Q/l^2 elementof_0 But E_2 = 0 That is the electric field inside the conducting plate = E_3 = 0 b Electric field lines are shown as 1 plate is negativity change and lower side positive charge with charge + Q d potential difference = D_v = E_1 d + E_2 x_0 + E_3 OR D_v = (E_1 + E_2) as = Q/l^2 elementof_0 (2d) \r\n So C = Q/D_v = Q/Q times 2d/l^2 elementof_0 = elementof_0 l^2/2d e Now there are two capacitors with capacitors c_1 and c_2 and both are in parallel. c_1 = elementof_0 lx/3d and c_2 = elementof_0 l(l - x)/2d and resultant c = c_1 + c_2 = Gol/d [x/3 + l - x/2] l = elementof_0 l(3l - x)/6d = elementof_0 l/d [3l - x/0]