Question

Two square metal plates of side 1 m are kept 0.01 m apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of emf 500 V. The plates are then lowered vertically into the oil at a speed of 0.001 m/s. Calculate the current drawn from the battery during the process. [Dielectric constant of oil = 11, $\epsilon_0 = 8.85 × 10^{-12} C^2N^{-1}m^{-2}$]

Answer 1

so 1st understand the normal approach

let it b (plate) lowered by x distance

so it is like 2 capasitors in parallel ..........Ct=E(1-x).1/d + kEx.1/d....C=EA/d

.... .Ct = (k-1)x + 1]E/d

........Qt/V = Ct........ Capcitance at itme t = ccharge at time t /voltage

....... ....... ... ... dQ/dt = [V.E.(k-1)/d.] (dx/dt)...... diffrentiating on both side and dividing by dt

......... I=V.E(k-1).v/d........... ..v is velocity ......ab kya put karo k= 11 v= .001 E=8.86 d=.01 V=500