Question

Each plate of a parallel-plate capacator is a square with side length r, and the plates are separated by a distance d. The capacitor is connected to a source of voltage, V. A plastic slab of thickness d and dielectric constant K is inserted slowly between the plates over the time period Δt until the slab is squarely between the plates. While the slab is being inserted, a current runs through the battery/capacitor circuit.

Assuming that the dielectric is inserted at a constant rate, find the current I as the slab is inserted.

Answer 1

`I=V\epsilon_o (K-1) * (r^2)/(d*\triangle t)`

Step-by-step explanation:

Since a slab is inserted between the plates, so we consider them as two capacitors attached in Parallel. One with dielectric and one without dielectric.

Equation will become:

Total Capacitance=Capacitance With dielectric+ Capacitance without Dielectric

`C=\epsilon_oK*a*(r)/(d)+ \epsilon_o(r-a)(r)/(d)`

Where:

a is the distance at which slab is added between the plates.

Rearranging the above equation:

`C=\epsilon_o*(r)/(d) [Ka+ (r-a)]`

Charge on the capacitor Q is given by:

Q=CV

Current "I" is given by:

`I=(dQ)/(dt)`

Now,

`I=(d(CV))/(dt) \\\\I=(Vd(C))/(dt) \\I=(Vd[\epsilon_o*(r)/(d) [Ka+ (r-a)]])/(dt) \\`

Taking derivative:

`I=V\epsilon_o}*\frac{r}d}[(d(Ka))/(dt)+(dr)/(dt)-(d(a))/(dt)]\\I=V\epsilon_o * (r)/(d) (K-1)(d(a))/(dt)` dr/dt=0 (r is constant)

In the above equation, d(a)/dt is the speed which is constant.

Speed= Distance/time

d(a)/dt= r/Δt

Final Equation will become:

`I=V\epsilon_o * (r)/(d) (K-1)(r)/(\triangle t) \\I=V\epsilon_o (K-1) * (r^2)/(d*\triangle t)`