Four short ones: geometry, slabs, and stored energy
Each rests on the single formula \(C=\dfrac{K\varepsilon_0 A}{d}\) — plus knowing what stays fixed when you change the capacitor: charge, or voltage.
An air parallel-plate condenser has capacity \(10^{-12}\,\text{F}\). The plates are moved apart so the separation doubles, and the gap is filled with a dielectric of constant \(K=4.0\). Find the new capacity (in farad).
- Original C₀10⁻¹² F
- New separation2d
- Dielectric K4.0
The new capacitance \(C'\).
Capacitance is \(C=\dfrac{K\varepsilon_0 A}{d}\). Two changes act here: doubling \(d\) halves C, while a dielectric \(K=4\) multiplies it by 4. Multiply the two factors together.
- Write the ratio. \(\dfrac{C'}{C_0}=\dfrac{K}{(d\to 2d)\text{ factor}}=\dfrac{4}{2}=2\).
- Apply. $$C'=2\times C_0 = 2\times 10^{-12}\,\text{F}$$
Stack the factors: ×K for the dielectric, ÷2 for the doubled gap. \(4 \div 2 = 2\), so the capacity simply doubles. The "4×" option is the trap if you forget the plates moved apart.
Three condensers, each \(C\,\mu\text{F}\), are connected in series. An exactly similar set is connected in parallel to the first. If the combination is \(4\,\mu\text{F}\), find \(C\).
- Each capC μF
- Each branch3 in series
- Branches2 in parallel
- Total4 μF
The value of \(C\).
Each branch is three equal caps in series → \(\dfrac{C}{3}\). Two such branches in parallel add: \(\dfrac{C}{3}+\dfrac{C}{3}=\dfrac{2C}{3}\). Set that equal to the known total.
- Each branch. \(\dfrac{C}{3}\).
- Two branches (parallel). \(\dfrac{2C}{3}\).
- Solve. $$\frac{2C}{3}=4 \;\Rightarrow\; C=6\,\mu\text{F}$$
Three-in-series then doubled = \(\tfrac{2C}{3}\). Just invert: \(C=\tfrac{3}{2}\times 4 = 6\). The "4" option is bait — that's the combination value, not C.
A parallel-plate capacitor has separation \(d\) and capacitance \(25\,\mu\text{F}\). A metallic foil of thickness \(\tfrac{2}{7}d\) is introduced between the plates. Find the new capacitance.
- Original C25 μF
- Separationd
- Foil thickness2d/7 (metal)
The new capacitance \(C'\) with the foil inserted.
A conductor (not a dielectric) carries no field inside it — it simply removes its own thickness from the effective gap, wherever it sits. So the formula becomes \(C'=\dfrac{\varepsilon_0 A}{d-t}\) with \(t=\tfrac{2}{7}d\).
- Effective gap. \(d-\dfrac{2}{7}d=\dfrac{5}{7}d\).
- Ratio of capacitances. \(\dfrac{C'}{C}=\dfrac{d}{\tfrac57 d}=\dfrac{7}{5}\).
- Apply. $$C'=\frac{7}{5}\times 25 = 35\,\mu\text{F}$$
Metal slab → subtract its thickness from d; position doesn't matter. Gap shrinks to \(\tfrac57 d\), so C grows by \(\tfrac75\): \(25\to 35\). (A dielectric would behave differently — only a conductor removes its full thickness.)
A charged parallel-plate condenser with air has capacity \(C_0\) and stored energy \(W_0\). The air is replaced by glass (\(K=5\)). Find the new capacity and stored energy.
- OriginalC₀, W₀ (air)
- Dielectric K5 (glass)
- StateCharged & isolated (Q fixed)
The new capacity and stored energy.
The condenser is charged then left alone, so the charge Q is fixed. Capacity always rises with the dielectric: \(C=KC_0=5C_0\). For energy at constant charge use \(W=\dfrac{Q^2}{2C}\) — since \(C\) goes up by 5, \(W\) goes down by 5.
- New capacity. \(C = KC_0 = 5C_0\).
- New energy (Q fixed). $$W=\frac{Q^2}{2C}=\frac{Q^2}{2(5C_0)}=\frac{W_0}{5}$$
Isolated capacitor + dielectric → C up by K, energy down by K. The field pulls the slab in and does the work, so stored energy drops: \(5C_0\) and \(\tfrac{W_0}{5}\). (If a battery had stayed connected, V is fixed instead and energy would rise to \(5W_0\) — that's option (a)'s scenario.)
At constant charge: \(C=5C_0\) and \(W=\dfrac{W_0}{5}\).