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 (potential energy of the square array):

Step 1 (Formula): The potential energy of a system of point charges is the sum over every distinct pair:
\[ U = \frac{1}{4\pi\varepsilon_0}\sum_{\text{pairs}} \frac{q_i q_j}{r_{ij}} \]
with \(k = \dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N m}^2/\text{C}^2\).

Step 2 (Label and data): Side \(a = 1\ \text{cm} = 0.01\ \text{m}\); each charge magnitude \(q = 1\ \mu C = 10^{-6}\ \text{C}\), so \(q^2 = 10^{-12}\ \text{C}^2\). Top corners \(+q, +q\); bottom corners \(-q, -q\). There are 6 pairs: 4 sides (length \(a\)) and 2 diagonals (length \(a\sqrt{2}\)).

Step 3 (Four side pairs): Products of charges along the sides are: top \((+q)(+q)=+q^2\); bottom \((-q)(-q)=+q^2\); left \((+q)(-q)=-q^2\); right \((+q)(-q)=-q^2\). Their sum \(= +q^2+q^2-q^2-q^2 = 0\). So the four sides contribute nothing:
\[ U_{\text{sides}} = \frac{k}{a}(0) = 0 \]

Step 4 (Two diagonal pairs): Each diagonal joins a \(+q\) to a \(-q\), giving product \(-q^2\), at distance \(a\sqrt{2}\):
\[ U_{\text{diag}} = \frac{k}{a\sqrt{2}}(-q^2 - q^2) = \frac{-2kq^2}{a\sqrt{2}} = -\frac{\sqrt{2}\,kq^2}{a} \]

Step 5 (Numbers): \(\dfrac{kq^2}{a} = \dfrac{9\times10^{9}\times10^{-12}}{0.01} = 0.9\ \text{J}\). Hence
\[ U = -\sqrt{2}\times 0.9 = -1.27\ \text{J} \]
\[\boxed{U \approx -1.27\ \text{J}}\]
The negative sign shows the system is bound (attractive overall).

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Option 2 (capacitor with dielectric slab):

Step 1 (Charge stored, battery still on): \(C_0 = 20\ \text{pF} = 20\times10^{-12}\ \text{F}\), \(V = 100\ \text{V}\).
\[ Q = C_0 V = 20\times10^{-12}\times100 = 2\times10^{-9}\ \text{C} = 2\ \text{nC} \]

Step 2 (Battery removed): With the battery disconnected the plates are isolated, so the charge cannot change. Therefore the final charge is the same:
\[ \boxed{Q = 2\times10^{-9}\ \text{C} = 2\ \text{nC}} \ \text{(part i)} \]

Step 3 (New capacitance with the slab): For a slab of thickness \(t\) and dielectric constant \(K\) inside a gap \(d\),
\[ C = \frac{\varepsilon_0 A}{\,d - t + \dfrac{t}{K}\,} \]
From \(C_0 = \varepsilon_0 A/d\), \(\varepsilon_0 A = C_0 d = 20\times10^{-12}\times4\times10^{-3} = 8\times10^{-14}\).

Step 4 (Effective gap): \(d - t + t/K = 4 - 2 + \dfrac{2}{4} = 2.5\ \text{mm} = 2.5\times10^{-3}\ \text{m}\). So
\[ C = \frac{8\times10^{-14}}{2.5\times10^{-3}} = 3.2\times10^{-11}\ \text{F} = 32\ \text{pF} \]

Step 5 (New potential difference):
\[ V' = \frac{Q}{C} = \frac{2\times10^{-9}}{3.2\times10^{-11}} = 62.5\ \text{V} \]
\[\boxed{V' = 62.5\ \text{V}} \ \text{(part ii)}\]