Chapter 2: Electrostatic Potential and Capacitance

Part (1): Potential at P

The potential at a point at distance $$r$$ from a point charge $$q$$ is $$V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$$.

Here $$q = 4 \times 10^{-7}\,\mathrm{C}$$, $$r = 9\,\mathrm{cm} = 9 \times 10^{-2}\,\mathrm{m}$$, and $$\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\,\mathrm{N\,m^2\,C^{-2}}$$.

$$V = \dfrac{(9 \times 10^{9})(4 \times 10^{-7})}{9 \times 10^{-2}}$$

$$V = \dfrac{3600}{9 \times 10^{-2}} = 4 \times 10^{4}\,\mathrm{V}$$

Part (2): Work done

The work done in bringing a charge $$q_0$$ from infinity (where the potential is zero) to a point where the potential is $$V$$ is $$W = q_0 V$$.

With $$q_0 = 2 \times 10^{-9}\,\mathrm{C}$$:

$$W = (2 \times 10^{-9})(4 \times 10^{4}) = 8 \times 10^{-5}\,\mathrm{J}$$

No, the work done does not depend on the path. The electrostatic force is conservative, so the work done depends only on the initial and final positions of the charge, not on the route taken to move it.

Let the charges be $$q_1 = 3 \times 10^{-8}\,\mathrm{C}$$ at point A and $$q_2 = -2 \times 10^{-8}\,\mathrm{C}$$ at point B, with $$\mathrm{AB} = 15\,\mathrm{cm}$$.

The total potential at a point is the algebraic sum of the potentials due to each charge. We look for points on the line AB where this sum is zero.

Case 1: Point P between the charges.

Let P be at a distance $$x$$ (in cm) from A, so it is $$(15 - x)$$ cm from B.

$$V = \dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q_1}{x} + \dfrac{q_2}{15-x}\right] = 0$$

$$\dfrac{3 \times 10^{-8}}{x} + \dfrac{-2 \times 10^{-8}}{15-x} = 0$$

$$\dfrac{3}{x} = \dfrac{2}{15-x}$$

$$3(15 - x) = 2x \quad\Rightarrow\quad 45 - 3x = 2x \quad\Rightarrow\quad 5x = 45 \quad\Rightarrow\quad x = 9\,\mathrm{cm}$$

So the potential is zero at a point 9 cm from the charge $$q_1$$ (i.e. 6 cm from $$q_2$$).

Case 2: Point P outside, beyond B.

Let P be at a distance $$x$$ (in cm) from A, beyond B, so it is $$(x - 15)$$ cm from B.

$$\dfrac{3 \times 10^{-8}}{x} + \dfrac{-2 \times 10^{-8}}{x-15} = 0$$

$$\dfrac{3}{x} = \dfrac{2}{x-15}$$

$$3(x - 15) = 2x \quad\Rightarrow\quad 3x - 45 = 2x \quad\Rightarrow\quad x = 45\,\mathrm{cm}$$

(On the side beyond A the larger positive charge always dominates, so there is no zero of potential there.)

Hence the potential is zero at two points on the line: 9 cm from $$q_1$$ (between the charges), and 45 cm from $$q_1$$ (i.e. 30 cm beyond $$q_2$$).

In Fig. 2.8(a) the field lines belong to a positive point charge, with P nearer the charge than Q. In Fig. 2.8(b) they belong to a negative point charge, with A nearer the charge than B.

(1) Signs of the potential differences

For a point charge, $$V \propto \dfrac{q}{r}$$.

Near a positive charge, $$V$$ is positive and decreases with distance. Since P is closer than Q, $$V_P > V_Q$$, so $$V_P - V_Q$$ is positive.

Near a negative charge, $$V$$ is negative and its magnitude decreases with distance. B is farther, so $$V_B$$ is less negative than $$V_A$$; thus $$V_B > V_A$$, and $$V_B - V_A$$ is positive.

(2) Sign of the potential energy difference of a small negative charge

The potential energy of a charge $$q$$ at a point of potential $$V$$ is $$U = qV$$. For a negative charge, $$q = -|q|$$.

Between Q and P: $$U_Q - U_P = -|q|(V_Q - V_P)$$. Since $$V_Q - V_P < 0$$, this gives $$U_Q - U_P > 0$$ — positive.

Between A and B: $$U_A - U_B = -|q|(V_A - V_B)$$. Since $$V_A - V_B < 0$$, this gives $$U_A - U_B > 0$$ — positive.

(3) Sign of the work done by the field in moving a small positive charge from Q to P

The work done by the field is $$W_{field} = q(V_Q - V_P)$$. With $$q > 0$$ and $$V_Q - V_P < 0$$, $$W_{field}$$ is negative. (To carry a positive charge from low to high potential, work must be done against the field.)

(4) Sign of the work done by the external agency in moving a small negative charge from B to A

The work done by an external agency (moving the charge slowly) is $$W_{ext} = q(V_A - V_B)$$. With $$q = -|q| < 0$$ and $$V_A - V_B < 0$$, the product is positive.

(5) Kinetic energy of a small negative charge going from B to A

From part (2), the potential energy of the negative charge is greater at A than at B ($$U_A > U_B$$). By conservation of energy, $$KE + U = \text{constant}$$, so as $$U$$ increases the kinetic energy must decrease. (Physically, the negative test charge is repelled by the negative source charge, so it slows down as it approaches A.)

From Fig. 2.15 the corners carry charges $$+q$$ at A, $$-q$$ at B, $$+q$$ at C and $$-q$$ at D (like charges at opposite corners).

Part (1): Work to put together this arrangement

The work required equals the total electrostatic potential energy of the configuration, which is the sum of the potential energies of every distinct pair of charges:

$$U_{pair} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_iq_j}{r_{ij}}$$

There are 6 pairs. The four sides AB, BC, CD, DA each have length $$d$$; the two diagonals AC and BD each have length $$d\sqrt{2}$$.

Sides — each side joins a $$+q$$ and a $$-q$$, so each contributes $$\dfrac{1}{4\pi\varepsilon_0}\dfrac{(+q)(-q)}{d} = -\dfrac{q^2}{4\pi\varepsilon_0 d}$$. The four sides together give $$-\dfrac{4q^2}{4\pi\varepsilon_0 d}$$.

Diagonals — AC joins $$+q$$ and $$+q$$; BD joins $$-q$$ and $$-q$$. Each contributes $$+\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{d\sqrt2}$$. The two diagonals together give $$+\dfrac{2q^2}{4\pi\varepsilon_0\,d\sqrt2}$$.

Adding all six pairs:

$$W = \dfrac{q^2}{4\pi\varepsilon_0 d}\left[-4 + \dfrac{2}{\sqrt2}\right] = \dfrac{q^2}{4\pi\varepsilon_0 d}\left[-4 + \sqrt2\right]$$

$$W = -\dfrac{q^2}{4\pi\varepsilon_0 d}\left(4 - \sqrt2\right)$$

The work is negative, which means the assembled configuration has lower energy than the fully separated charges; energy is in fact released in assembling it. (Putting the charges together in any other order gives the same total, since electrostatic energy is independent of the order of assembly.)

Part (2): Extra work to bring $$q_0$$ to the centre E

The extra work needed is $$W' = q_0 V_E$$, where $$V_E$$ is the potential at the centre E due to the four corner charges.

The centre of the square is equidistant from all four corners; that distance is half the diagonal, $$r = \dfrac{d\sqrt2}{2} = \dfrac{d}{\sqrt2}$$.

$$V_E = \dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{+q}{r} + \dfrac{-q}{r} + \dfrac{+q}{r} + \dfrac{-q}{r}\right] = 0$$

Since the two $$+q$$ and two $$-q$$ charges are all the same distance from E, the potential due to A and C is exactly cancelled by that due to B and D. Therefore

$$W' = q_0 V_E = 0$$

No extra work is needed to bring the charge $$q_0$$ to the centre.

Part (1): Potential energy of the two-charge system

The two charges are $$q_1 = 7\,\mathrm{\mu C} = 7 \times 10^{-6}\,\mathrm{C}$$ and $$q_2 = -2\,\mathrm{\mu C} = -2 \times 10^{-6}\,\mathrm{C}$$, placed at $$(-9\,\mathrm{cm},0,0)$$ and $$(9\,\mathrm{cm},0,0)$$. Their separation is $$r = 18\,\mathrm{cm} = 0.18\,\mathrm{m}$$.

$$U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r} = (9 \times 10^{9})\dfrac{(7 \times 10^{-6})(-2 \times 10^{-6})}{0.18}$$

$$U = (9 \times 10^{9})\dfrac{-14 \times 10^{-12}}{0.18} = -0.7\,\mathrm{J}$$

Part (2): Work to separate the charges infinitely

When the charges are infinitely far apart, the potential energy is zero. The work done by an external agency equals the change in potential energy:

$$W = U_{final} - U_{initial} = 0 - (-0.7\,\mathrm{J}) = 0.7\,\mathrm{J}$$

Part (3): Electrostatic energy in the external field

The external field is $$E = A/r^2$$ with $$A = 9 \times 10^{5}\,\mathrm{N\,C^{-1}\,m^2}$$. The corresponding external potential (that of a point-type source) is $$V(r) = A/r$$.

Each charge lies at a distance $$r = 9\,\mathrm{cm} = 0.09\,\mathrm{m}$$ from the origin, so the external potential at the location of each charge is

$$V = \dfrac{A}{r} = \dfrac{9 \times 10^{5}}{0.09} = 1 \times 10^{7}\,\mathrm{V}$$

The energy of the two charges in the external field is

$$U_{ext} = q_1V + q_2V = (7 \times 10^{-6})(10^{7}) + (-2 \times 10^{-6})(10^{7})$$

$$U_{ext} = 70 - 20 = 50\,\mathrm{J}$$

The total electrostatic energy is the sum of this and the mutual interaction energy of the pair (unchanged from Part 1):

$$U_{total} = U_{ext} + U_{mutual} = 50\,\mathrm{J} + (-0.7\,\mathrm{J}) = 49.3\,\mathrm{J}$$

Each molecule has a dipole moment $$p = 10^{-29}\,\mathrm{C\,m}$$. One mole contains Avogadro's number $$N = 6.023 \times 10^{23}$$ molecules. The field magnitude is $$E = 10^{6}\,\mathrm{V\,m^{-1}}$$.

The potential energy of a dipole making angle $$\theta$$ with the field is $$U(\theta) = -pE\cos\theta$$.

Initially, with 100% polarisation, every dipole is aligned with the field: $$\theta = 0$$, so the energy per dipole is $$U_i = -pE$$.

When the field direction is suddenly turned through $$60^\circ$$, each dipole momentarily makes an angle of $$60^\circ$$ with the new field, with energy $$U = -pE\cos 60^\circ$$. The dipoles then swing round to line up with the new direction ($$\theta = 0$$ again).

The heat released per dipole equals the drop in its potential energy as it realigns:

$$\Delta U = \big(-pE\cos 60^\circ\big) - \big(-pE\big) = pE\,(1 - \cos 60^\circ)$$

$$\Delta U = pE\left(1 - \tfrac{1}{2}\right) = \tfrac{1}{2}pE$$

For one mole (all $$N$$ dipoles), the total heat released is

$$Q = \tfrac{1}{2}\,NpE = \tfrac{1}{2}\,(6.023 \times 10^{23})(10^{-29})(10^{6})$$

$$Q = \tfrac{1}{2}\,(6.023 \times 10^{\,23-29+6}) = \tfrac{1}{2}\,(6.023) \approx 3\,\mathrm{J}$$

So the substance releases about 3 J of heat in re-aligning its dipoles along the new field direction.

(1) Comb attracting bits of paper

Running a comb through dry hair charges the comb by friction. When the charged comb is brought near small bits of paper, it polarises them: the molecules of the (non-conducting) paper become induced dipoles, with the side facing the comb acquiring the opposite charge. Because the induced opposite charge is nearer to the comb than the induced like charge, the net force is one of attraction, and the bits of paper are picked up.

If the hair is wet, or on a rainy day, the moisture makes the hair (and the surrounding air) conducting. The charge produced on the comb leaks away through the moist hair, so the comb does not stay charged and cannot attract the paper.

(2) Slightly conducting aircraft tyres

As an aircraft moves, and especially during landing, friction with the air builds up a large amount of static charge on it. If this charge could not escape, the resulting high potential could produce a spark, which is dangerous near the fuel. Making the tyres slightly conducting provides a path for the accumulated charge to flow harmlessly to the ground, preventing such sparking.

(3) Metallic ropes on vehicles carrying inflammable material

The motion of the vehicle generates static charge by friction. If this charge were allowed to accumulate, a spark could occur and ignite the inflammable cargo. A metallic rope touching the ground continuously conducts the charge away to the earth, keeping the vehicle uncharged and preventing any spark.

(4) Bird on a power line versus a man touching it

A bird perched on a single bare wire touches the line at only one point; both its feet are at essentially the same potential, so there is no potential difference across its body and hence no current flows through it. A man standing on the ground who touches the line completes a conducting path between the high-potential line and the earth (at zero potential). The large potential difference drives a current through his body to the ground, giving a fatal shock.

Let the plate area be $$A$$ and the plate separation be $$d$$. Without the slab, the capacitance is

$$C_0 = \dfrac{\varepsilon_0 A}{d}$$

The slab has dielectric constant $$K$$ and thickness $$t = \tfrac{3}{4}d$$. The remaining gap is air, of thickness $$d - t = d - \tfrac{3}{4}d = \tfrac{1}{4}d$$.

The arrangement behaves as two capacitors in series: the air gap and the dielectric slab, since the same charge resides on the plates and the potential differences add.

Air-gap capacitor (thickness $$\tfrac{1}{4}d$$):

$$C_1 = \dfrac{\varepsilon_0 A}{d/4} = \dfrac{4\varepsilon_0 A}{d}$$

Dielectric-slab capacitor (thickness $$\tfrac{3}{4}d$$):

$$C_2 = \dfrac{K\varepsilon_0 A}{3d/4} = \dfrac{4K\varepsilon_0 A}{3d}$$

For a series combination,

$$\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{d}{4\varepsilon_0 A} + \dfrac{3d}{4K\varepsilon_0 A}$$

$$\dfrac{1}{C} = \dfrac{d}{4\varepsilon_0 A}\left(1 + \dfrac{3}{K}\right) = \dfrac{d}{4\varepsilon_0 A}\cdot\dfrac{K + 3}{K}$$

$$C = \dfrac{4K}{K + 3}\cdot\dfrac{\varepsilon_0 A}{d} = \dfrac{4K}{K + 3}\,C_0$$

Since $$K > 1$$, the factor $$\dfrac{4K}{K+3}$$ is greater than 1, so inserting the slab increases the capacitance. (If $$K$$ is very large, $$C \to 4C_0$$, i.e. the capacitance approaches that of a plate separation of only $$d/4$$ — the air gap then sets the limit.)

In the network of Fig. 2.29 the capacitors $$C_2$$, $$C_3$$ and $$C_4$$ are connected in series, and this series combination is in parallel with $$C_1$$ across the 500 V supply. Each capacitor has $$C = 10\,\mathrm{\mu F}$$.

Part (1): Equivalent capacitance of the network

For $$C_2$$, $$C_3$$, $$C_4$$ in series, the effective capacitance $$C'$$ is given by

$$\dfrac{1}{C'} = \dfrac{1}{C_2} + \dfrac{1}{C_3} + \dfrac{1}{C_4} = \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10} = \dfrac{3}{10}$$

$$C' = \dfrac{10}{3}\,\mathrm{\mu F}$$

This series combination is in parallel with $$C_1$$, so the equivalent capacitance of the network is

$$C = C' + C_1 = \dfrac{10}{3} + 10 = \dfrac{40}{3} \approx 13.3\,\mathrm{\mu F}$$

Part (2): Charge on each capacitor

$$C_1$$ is connected directly across the 500 V supply, so the full 500 V appears across it:

$$Q_1 = C_1 V = (10 \times 10^{-6})(500) = 5 \times 10^{-3}\,\mathrm{C}$$

The series branch $$C_2C_3C_4$$ also has 500 V across it. In a series combination every capacitor carries the same charge, equal to that of the equivalent capacitor $$C'$$:

$$Q_2 = Q_3 = Q_4 = C'V = \left(\dfrac{10}{3} \times 10^{-6}\right)(500) \approx 1.7 \times 10^{-3}\,\mathrm{C}$$

The potential difference across each of $$C_2$$, $$C_3$$, $$C_4$$ is

$$V_2 = V_3 = V_4 = \dfrac{Q_2}{C} = \dfrac{1.7 \times 10^{-3}}{10 \times 10^{-6}} \approx 167\,\mathrm{V}$$

(As a check, $$167 \times 3 \approx 500\,\mathrm{V}$$, the voltage across the series branch.)

Part (1): Energy stored in the single capacitor

The capacitor has $$C = 900\,\mathrm{pF} = 900 \times 10^{-12}\,\mathrm{F}$$ and is charged to $$V = 100\,\mathrm{V}$$.

The energy stored is

$$U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(900 \times 10^{-12})(100)^2$$

$$U = \tfrac{1}{2}(900 \times 10^{-12})(10^{4}) = 4.5 \times 10^{-6}\,\mathrm{J}$$

Part (2): Energy after connecting to the second capacitor

The charge initially on the capacitor is

$$Q = CV = (900 \times 10^{-12})(100) = 9 \times 10^{-8}\,\mathrm{C}$$

When it is disconnected from the battery and connected to an identical uncharged 900 pF capacitor, the total charge $$Q$$ is conserved and is shared by the two capacitors, which are now in parallel (total capacitance $$2C = 1800\,\mathrm{pF}$$). They settle to a common potential

$$V' = \dfrac{Q}{2C} = \dfrac{9 \times 10^{-8}}{1800 \times 10^{-12}} = 50\,\mathrm{V}$$

The energy stored by the system is now

$$U' = \tfrac{1}{2}(2C)V'^2 = \tfrac{1}{2}(1800 \times 10^{-12})(50)^2$$

$$U' = \tfrac{1}{2}(1800 \times 10^{-12})(2500) = 2.25 \times 10^{-6}\,\mathrm{J}$$

The final energy ($$2.25\,\mathrm{\mu J}$$) is only half the initial energy ($$4.5\,\mathrm{\mu J}$$). The 'missing' energy, $$2.25 \times 10^{-6}\,\mathrm{J}$$, is dissipated as heat (and a little electromagnetic radiation) in the connecting wires while the charge redistributes.

Let $$q_1 = 5 \times 10^{-8}\,\mathrm{C}$$ be at point A and $$q_2 = -3 \times 10^{-8}\,\mathrm{C}$$ be at point B, with $$\mathrm{AB} = 16\,\mathrm{cm}$$.

The net potential at a point is the algebraic sum of the potentials of the two charges. We find where this sum vanishes.

Case 1: Point P between the charges.

Let P be at a distance $$x$$ (in cm) from A, hence $$(16 - x)$$ cm from B.

$$\dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q_1}{x} + \dfrac{q_2}{16-x}\right] = 0$$

$$\dfrac{5 \times 10^{-8}}{x} + \dfrac{-3 \times 10^{-8}}{16-x} = 0$$

$$\dfrac{5}{x} = \dfrac{3}{16-x}$$

$$5(16 - x) = 3x \quad\Rightarrow\quad 80 - 5x = 3x \quad\Rightarrow\quad 8x = 80 \quad\Rightarrow\quad x = 10\,\mathrm{cm}$$

So the potential is zero at a point 10 cm from $$q_1$$ (and 6 cm from $$q_2$$).

Case 2: Point P outside, beyond B.

Let P be at a distance $$x$$ (in cm) from A, beyond B, hence $$(x - 16)$$ cm from B.

$$\dfrac{5 \times 10^{-8}}{x} + \dfrac{-3 \times 10^{-8}}{x-16} = 0$$

$$\dfrac{5}{x} = \dfrac{3}{x-16}$$

$$5(x - 16) = 3x \quad\Rightarrow\quad 5x - 80 = 3x \quad\Rightarrow\quad 2x = 80 \quad\Rightarrow\quad x = 40\,\mathrm{cm}$$

(On the side beyond A the positive charge $$q_1$$ always dominates, so there is no zero of potential there.)

Hence the potential is zero at two points on the line: 10 cm from $$q_1$$ (between the charges) and 40 cm from $$q_1$$ (i.e. 24 cm beyond $$q_2$$).

In a regular hexagon, the distance from the centre O to each vertex is equal to the length of a side. So each charge is at a distance

$$r = 10\,\mathrm{cm} = 0.10\,\mathrm{m}$$

from the centre.

There are 6 identical charges, each $$q = 5\,\mathrm{\mu C} = 5 \times 10^{-6}\,\mathrm{C}$$. Electric potential is a scalar, so the potential at the centre is simply the sum of the six equal contributions:

$$V = 6 \times \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$$

$$V = 6 \times (9 \times 10^{9}) \times \dfrac{5 \times 10^{-6}}{0.10}$$

$$V = 6 \times (9 \times 10^{9}) \times (5 \times 10^{-5})$$

$$V = 6 \times (4.5 \times 10^{5}) = 2.7 \times 10^{6}\,\mathrm{V}$$

For a parallel-plate capacitor with air between the plates,

$$C_0 = \dfrac{\varepsilon_0 A}{d} = 8\,\mathrm{pF}$$

where $$A$$ is the plate area and $$d$$ the separation.

Now the separation is reduced to $$d' = \dfrac{d}{2}$$ and the gap is filled with a dielectric of constant $$K = 6$$. The new capacitance is

$$C' = \dfrac{K\varepsilon_0 A}{d'} = \dfrac{K\varepsilon_0 A}{d/2} = \dfrac{2K\varepsilon_0 A}{d}$$

$$C' = 2K\left(\dfrac{\varepsilon_0 A}{d}\right) = 2K\,C_0$$

Substituting $$K = 6$$ and $$C_0 = 8\,\mathrm{pF}$$:

$$C' = 2 \times 6 \times 8\,\mathrm{pF} = 96\,\mathrm{pF}$$

The capacitance of a parallel-plate capacitor with air between the plates is

$$C = \dfrac{\varepsilon_0 A}{d}$$

with $$\varepsilon_0 = 8.85 \times 10^{-12}\,\mathrm{F\,m^{-1}}$$, plate area $$A = 6 \times 10^{-3}\,\mathrm{m^2}$$ and separation $$d = 3\,\mathrm{mm} = 3 \times 10^{-3}\,\mathrm{m}$$.

$$C = \dfrac{(8.85 \times 10^{-12})(6 \times 10^{-3})}{3 \times 10^{-3}}$$

$$C = (8.85 \times 10^{-12}) \times 2 = 1.77 \times 10^{-11}\,\mathrm{F} \approx 18\,\mathrm{pF}$$

When this capacitor is connected to a $$V = 100\,\mathrm{V}$$ supply, the charge on each plate is

$$Q = CV = (1.77 \times 10^{-11})(100)$$

$$Q = 1.77 \times 10^{-9}\,\mathrm{C} \approx 1.8 \times 10^{-9}\,\mathrm{C}$$

The energy stored in a capacitor charged to a potential difference $$V$$ is

$$U = \tfrac{1}{2}CV^2$$

Here $$C = 12\,\mathrm{pF} = 12 \times 10^{-12}\,\mathrm{F}$$ and $$V = 50\,\mathrm{V}$$.

$$U = \tfrac{1}{2}(12 \times 10^{-12})(50)^2$$

$$U = \tfrac{1}{2}(12 \times 10^{-12})(2500)$$

$$U = \tfrac{1}{2}(3 \times 10^{-8}) = 1.5 \times 10^{-8}\,\mathrm{J}$$

Initial state. The 600 pF capacitor is charged by a 200 V supply. Its charge and stored energy are

$$Q = CV = (600 \times 10^{-12})(200) = 1.2 \times 10^{-7}\,\mathrm{C}$$

$$U_i = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(600 \times 10^{-12})(200)^2$$

$$U_i = \tfrac{1}{2}(600 \times 10^{-12})(4 \times 10^{4}) = 1.2 \times 10^{-5}\,\mathrm{J}$$

After connecting to the uncharged capacitor. The capacitor is disconnected from the supply and connected to an identical uncharged 600 pF capacitor. The total charge $$Q$$ is conserved and is shared between the two capacitors, which are now in parallel (total capacitance $$2C = 1200\,\mathrm{pF}$$). They reach a common potential

$$V' = \dfrac{Q}{2C} = \dfrac{1.2 \times 10^{-7}}{1200 \times 10^{-12}} = 100\,\mathrm{V}$$

The final stored energy is

$$U_f = \tfrac{1}{2}(2C)V'^2 = \tfrac{1}{2}(1200 \times 10^{-12})(100)^2$$

$$U_f = \tfrac{1}{2}(1200 \times 10^{-12})(10^{4}) = 0.6 \times 10^{-5}\,\mathrm{J}$$

Energy lost.

$$\Delta U = U_i - U_f = 1.2 \times 10^{-5} - 0.6 \times 10^{-5} = 6 \times 10^{-6}\,\mathrm{J}$$

This energy, $$6 \times 10^{-6}\,\mathrm{J}$$, is dissipated as heat (and a small amount of electromagnetic radiation) in the connecting wires as the charge redistributes between the two capacitors.