Dielectric field & charge sharing — worked revision

Q31 · Q32
Q · 31

Field & potential across a centred slab

Question

A dielectric slab of thickness \(d\) is inserted in a parallel plate capacitor whose negative plate is at \(x = 0\) and positive plate is at \(x = 3d\). The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to \(3d\)
  • a) the magnitude of the electric field remains the same
  • b) the direction of the electric field remains the same
  • c) the electric potential increases continuously
  • d) the electric potential increases at first, then decreases and again increases

Multiple-correct: both (b) and (c) hold.

The Picture

Field across the gap

− + x=0 x=3d slab K d 2d E same direction, smaller in slab

Potential vs position

x V 0d 2d3d monotonic rise (gentler in slab)

Charge given ⇒ free charge (and so the field direction) is fixed: \(E\) points from \(+\) plate toward \(-\) plate everywhere. Inside the slab \(E\) is smaller \((E/K)\) but still the same direction, so \(V\) keeps rising — just less steeply.

Why each option

(a) |E| sameNo — \(E_{slab}=E_{air}/K\)
(b) direction sameYes — always \(+\!\to\!-\)
(c) V rises continuouslyYes — monotonic
(d) V up–down–upNo — never falls

Concept

With charge fixed, \(D=\sigma\) is uniform, so \(E=D/(\varepsilon_0\varepsilon_r)\) only changes magnitude where \(\varepsilon_r\) changes — never direction. Potential is \(V=-\int E\,dx\) walked from the \(-\) plate; since \(E\) keeps one sign, \(V\) climbs the whole way.

Easy Trick

A dielectric slab only weakens the field, it never reverses it. So direction is unchanged and potential marches steadily uphill from \(-\) to \(+\) plate — the V–x graph is steep-gentle-steep but always rising. Both (b) and (c).

b, c
Direction of \(E\) unchanged; potential rises continuously.
Q · 32

Charge sharing between two capacitors

Question

A 2 µF condenser is charged upto 200 V and then battery is removed. On combining this with another uncharged condenser in parallel, the potential differences between two plates are found to be 40 V. The capacity of second condenser is
  • a) 2 µF
  • b) 4 µF
  • c) 8 µF
  • d) 16 µF

The Picture — before & after

2 µF 200 V +Q −Q charged, isolated connect ∥ 2µF C₂ 40 V same charge, shared

Battery removed ⇒ the total charge is fixed. Connecting the uncharged \(C_2\) in parallel just lets that fixed charge spread over a larger capacitance, dropping the voltage from 200 V to 40 V.

Concept

Isolated ⇒ charge conserved: \(Q = C_1V_i = (C_1+C_2)V_f\). Solve for \(C_2\) from the voltage drop.

Method & Steps

  1. \(Q = 2\times200 = 400\ \mu C\)
  2. \(V_f=\dfrac{Q}{C_1+C_2}\Rightarrow 40=\dfrac{400}{2+C_2}\)
  3. \(2+C_2 = 10\)
  4. \(C_2 = 8\ \mu F\)

Easy Trick

Voltage fell \(200\to40\), a factor of 5, so the total capacitance must have grown 5×: \(C_1+C_2 = 5C_1 = 10\ \mu F\Rightarrow C_2 = 8\ \mu F\). (Charge stays put; only \(C\) and \(V\) trade off.)

c
\(C_2 = 8\ \mu F\)