Question

Compute the electric potential energy of the system with the help of the figure given above. The square has side 1 cm with charges \(+1\ \mu C\) and \(+1\ \mu C\) on the two top corners and \(-1\ \mu C\) and \(-1\ \mu C\) on the two bottom corners.
OR
The capacitance of a parallel plate capacitor is 20 pF and distance between the plates is 4 mm. It is charged by a battery of 100 volt and then battery is removed. Now a dielectric slab (K = 4) and 2 mm thickness is placed in between the plates. Find out: (i) the final charge on each plate (ii) the potential difference between the plates.

Hint:

Option 1: sum \(kq_iq_j/r\) over 6 pairs; the four edges cancel, only the two unlike diagonals remain. Option 2: charge is fixed after the battery is removed; use \(C = \varepsilon_0 A/(d-t+t/K)\) then \(V=Q/C\).

Answer 1

Option 1 (square array by grouping symmetry):

Step 1 (Bond-counting): Picture the four charges as forming bonds: 4 edge bonds of length \(a\) and 2 diagonal bonds of length \(a\sqrt{2}\). Each bond stores energy \(k q_i q_j / r\). Sum all bonds.

Step 2 (Edges cancel by pairing): The top edge \((+,+)\) is repulsive \((+kq^2/a)\) and the bottom edge \((-,-)\) is also repulsive \((+kq^2/a)\); the two vertical edges are each \((+,-)\) attractive \((-kq^2/a)\). Adding, \(+kq^2/a + kq^2/a - kq^2/a - kq^2/a = 0\). The edge energy vanishes.

Step 3 (Only diagonals survive): Both diagonals connect unlike charges, so each stores \(-kq^2/(a\sqrt{2})\). Two of them give
\[ U = -\frac{2kq^2}{a\sqrt2} = -\frac{\sqrt2\,kq^2}{a} \]

Step 4 (Plug numbers): With \(k=9\times10^{9}\), \(q^2=10^{-12}\), \(a=0.01\): \(kq^2/a = 0.9\ \text{J}\), so \(U = -1.414\times0.9 = -1.27\ \text{J}\).
\[\boxed{U \approx -1.27\ \text{J}}\]

---

Option 2 (capacitor via field/thickness reasoning):

Step 1 (Fix the charge first): Charged to \(V=100\) V on \(C_0 = 20\) pF, the plates hold \(Q = C_0 V = 2\ \text{nC}\). Once the battery is pulled off, the isolated plates keep this charge no matter what is inserted. Part (i): \(Q = 2\ \text{nC}\).

Step 2 (Split the gap into two regions): The 4 mm gap becomes a 2 mm air part in series with a 2 mm dielectric part. The equivalent capacitance of series regions gives
\[ C = \frac{\varepsilon_0 A}{(d-t) + t/K}. \]

Step 3 (Get \(\varepsilon_0 A\)): From \(C_0 = \varepsilon_0 A/d\), \(\varepsilon_0 A = 20\ \text{pF}\times 4\ \text{mm}\). Effective thickness \(= (4-2) + 2/4 = 2.5\ \text{mm}\), i.e. \(\tfrac{4}{2.5}\) smaller denominator, so \(C = 20\times\tfrac{4}{2.5} = 32\ \text{pF}\).

Step 4 (Voltage from \(Q=CV\)):
\[ V' = \frac{Q}{C} = \frac{2\times10^{-9}}{32\times10^{-12}} = 62.5\ \text{V} \]
The voltage drops from 100 V to 62.5 V because inserting the dielectric raises the capacitance while charge is fixed.
\[\boxed{Q = 2\ \text{nC},\quad V' = 62.5\ \text{V}}\]