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Some of the most important Physics Topics include energy, motion, and force.
How do you find the expression for potential energy of a charged conductor and Loss of Energy due to Sharing of Charge?
A certain amount of work has to be done in order to charge a conductor. The energy spent for doing that work remains stored in the charged conductor as potential energy. Essentially, the electric field of the conductor stores this energy.
Calculation: Let a conductor be charged with Q and let its capacitance he C. The potential of the conductor is V.
During charging we assume that the whole amount of charge is not given to the conductor at a time, rather it is charged gradually. At first the charge of the conductor is zero and so its potential is also zero. Gradually its potential increases due to accumulation of charges. So at the time of charging the conductor has no particular potential. Its potential becomes V when its charge is Q.
Initial potential = 0; final potential = V
Average of these potentials = \(\frac{0+V}{2}\) = \(\frac{V}{2}\)
Therefore, work done= average potential × charge
= \(\frac{V}{2}\) × Q = \(\frac{1}{2}\)QV = \(\frac{1}{2}\)CV ᐧ V [∵ Q = CV]
= \(\frac{1}{2}\)CV2 = \(\frac{1}{2}\)C × \(\left(\frac{Q}{C}\right)^2\) = \(\frac{1}{2} \cdot \frac{Q^2}{C}\)
This work is stored in the charged conductor as potential energy.
∴ Potential energy of a charged conducwr
= \(\frac{1}{2} Q V\) = \(\frac{1}{2} C V^2\) = \(\frac{1}{2} \frac{Q^2}{C}\)
If C, V and Q are expressed in esu, the unit of potential energy will be erg. Again if C, V and Q are expressed in farad, volt and coulomb, respectively, the unit of potential energy will be joule.
Derivation using calculus: Let at any moment the charge on the conductor be q and its potential be v.
Evidently, q = Cν
When a charge dq is given to the conductor the work done against the repulsive force due to potential u is given by, dW = vdq = \(\frac{q}{C}\)dq
Hence the total work done to impart Q amount of charge is,
This work is stored as potential energy in the charged conductor.
Distribution Of Charge Between Two Conductors
Two conductors at the same potential: Let us consider two insulated uncharged conductors A and B of capacitances C1 and C2, respectively. They are connected by a fine metal wire [Fig]. Under this condition, if a charge Q be
given to this combination, it will he distributed between the two
conductors. Let us consider that the conductor A has obtained a charge Q1 and the conductor B a charge Q2. As the two conductors are connected to each other, they have the same potential. Let their common potential be V.
Therefore, charge on each conductor is proportional to its capacitance. If the two conductors have the same capacitances the given charge will be shared equally between them. If they have radii r1 cm and r2 cm, then C1 = r1 and C2 = r2 (in CGS system). In that case,
Therefore, charge on each spherical conductor is proportional to its radius.
We know that the surface density of charge on the surface of a charged sphere is the same everywhere. If σ1 and σ2 be the surface densities of charge of the two conductors, then
Therefore, surface density of charge of a spherical conductor is inversely proportional to its radius.
Two conductors initially at different potentials : Let us consider Iwo insulated conductors A and B. They have capacitances C1 and C2 and they are given charges Q1 and Q2 separately. So under this condition,
potential of the conductor A, V1 = \(\frac{Q_1}{C_1}\) or, Q1 = C1 V1
and potential of the conductor B, V2 = \(\frac{Q_2}{C_2}\) or, Q2 = C2V2
∴ Total charge of the conductors,
Q = Q1 + Q2 = C1V1 + C2V2 ……… (6)
If now the two conductors are connected by a thin metal wire, positive charge will flow from the conductor at higher potential to that at lower potential and this flow of charge will continue till their potentials become equal.
Suppose, V1 > V2; then charge will flow from A to B. Iet V he
the common potential after connection. During this flow of charge, total charge of the system remains constant.
So, total charge before connection = total charge after connection
i.e., Q = C1V1 + C2V2 = C1V + C2V = (C1 + C2)V
or, V = \(\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\) ……. (7)
After connection, if A and B contain charges q1 and q2 respectively, then
From equations (9) and (10) we get, \(q_1^{\prime}\) = \(q_2^{\prime}\), ie., the charge gained by the conductor B is equal to that lost by the conductor
Loss of energy due to sharing of charge:
Before connection, total energy of the two conductors
Now, C1 and C2 are both positive quantities and (V1 – V1)2, being a perfect square, is also positive. So, the relation (11) is positive.
Hence, there is always a loss of energy in the electric field of the conductors due to sharing of charges. According to the law of conservation of energy, this loss of energy must be converted to some other form, usually as heat in the connecting wire. This loss is partly converted into light and sound in addition to heat if sparkling occurs.
Numerical Examples
Example 1.
A conductor of capacity 4units, charged with 100 units of positive charge is connected to another conductor of capacity 2 units, charged with 20 units of negative charge. What is the change in potential of each conductor? What will be the charges of each of them after connection?
Solution:
Capacity of the first conductor, C1 = -4 units; charge, Q1 = 100 units.
∴ Potential, V1 = \(\frac{Q_1}{C_1}\) = \(\frac{100}{4}\) = 25 units
Capacity of the second conductor, C2 = 2 units; charge, Q2 = -20 units.
∴ Potential, V2 = \(\frac{Q_2}{C_2}\) = \(\frac{-20}{2}\) = -10 units
After connection, suppose, the common potential of the two conductors becomes equal to V.
∴ V = \(\frac{Q_1+Q_2}{C_1+C_2}\) = \(\frac{100-20}{4+2}\) = \(\frac{80}{6}\) = \(\frac{40}{3}\) = 13.33 units
So, the change of potential of first conductor
= 25- 13.33 = 11.67 units
The change of potential of second conductor
= 13.33 – (-10) = 23.33 units
Residual charge in the first conductor after connection,
q1 = C1V = 4 × \(\frac{40}{3}\) = 53.33 units
Residual charge in the second conductor after connection,
q2 = C2V = 2 × \(\frac{40}{3}\) = 26.67 units
Example 2.
An insulated metallic vessel full of water is charged with a potential of 3V. Drops of water are trickling from an orifice at the bottom of the vessel. What is the amount of charge contained in each spherical drop of radius 1mm?
Solution:
Potential of the metal vessel full of water, V = 3V
Radius of a water drop, R = 1 mm = 10-3 m
∴ Capacitance of the water drop,
C = 4πε0 R = \(\frac{10^{-3}}{9 \times 10^9}\) = \(\frac{1}{9} \times 10^{-12} \mathrm{~F}\)
∴ Charge of each water drop,
Q = CV = \(\frac{1}{9}\) × 10-12 × 3 = 3.3 × 10-13 C
Example 3.
The radii of two insulated metal spheres are 5 cm and 10 cm. They are charged up to potentials of 10 esu and 15 esu, respectively. If the two spheres are connected with one another, what will be the loss of energy?
Solution:
Radius of the first sphere = 5 cm
∴ Capacitance, C1 = 5 statF
and potential, V1 = 10 statV
∴ Charge, Q1 = C1 V1 = 5 × 10 = 50 statC
Radius of the second sphere = 10 cm
∴ Capacitance, C2 = 10 statF
and potential, V2 = 15 statV
∴ Charge, Q2 = C2V2 = 10 × 15 = 150 statC
Total charge of the two spheres,
Q = Q1 + Q2 = 50 + 150 = 200 statC
Equivalent capacitance of the combination of two spheres,
C = C1 + C2 = 5 + 10 = 15 statF
If V be the common potential of the two spheres after connection, then
Total energy of the two spheres after connections
E2 = \(\frac{1}{2} C V^2\) = \(\frac{1}{2} \times 15 \times\left(\frac{40}{3}\right)^2\) = 1333.33 erg
∴ Loss of energy due to connection
= 1375 – 1333.33 = 41.67 erg
Example 4.
A metal sphere of radius 10 cm is charged up to a poential of 80 esu. After sharing its charge with another sphere, the common potential of them becomes 20 esu. Wha is the radius of the second sphere?
Solution:
Radius of the first sphere 10 cm
∴ Capacitance C1 = 10 statF
and potential, V1 = 80 statV
∴ Charge, Q1 = C1 V1 = 10 × 80 = 800 statC
After connection, the common potential, V = 20 statV
Example 5.
One thousand similar electrified raindrops merge Into a single one so that their total charge remains unchanged. Find the change in the total electrostatic energy of the drops, assuming that all the drops are spherical and the small drops were initially at large distances from one another.
Solution:
Suppose, each drop of radius r contains a charge Q.
∴ Capacitance, C = r
and energy, E1 = \(\frac{1}{2} \cdot \frac{Q^2}{C}=\frac{Q^2}{2 r}\)
∴ Total energy of 1000 drops,
E = 1000E1 = \(\frac{500 Q^2}{r}\)
If R be the radius of the large drop then,
\(\frac{4}{3} \pi R^3\) = 1000 × \(\frac{4}{3} \pi r^3\) or, R = 10r
Charge of the large drop = 1000Q
∴ Energy of the large drop,
E2 = \(\frac{1}{2} \cdot \frac{(1000 Q)^2}{R}\) = \(\frac{5 \times 10^5 Q^2}{10 r}\) = \(\frac{5 \times 10^4 Q^2}{r}\)
∴ \(\frac{E_2}{E}\) = \(\frac{5 \times 10^4}{500}\) = 100 or, E2 = 100E
∴ Change in electrostatic energy of the drops
E2 – E = 100E – E = 99E = 99 × initial energy
Example 6.
Two equally charged soap bubbles of equal volume join together to form a large bubble. If each small bubble had a potential V, find the potential of the resultant bubble. [WBJEE 2000]
Soluton:
Let radius of each small bubble be r, charge be Q and radius of large bubble be R.
∴ \(\frac{4}{3} \pi R^3\) = 2 × \(\frac{4}{3} \pi r^3\) or, R = 21/3 ᐧ r
Potential of each small bubble,
V = \(\frac{Q}{r}\) or, Q = Vr
∴ Potential of the large bubble \(=\frac{\text { total charge }}{\text { radius }}\)
= \(\frac{2 Q}{R}\) = \(\frac{2 V r}{2^{1 / 3} r}\) = 22/3V
Example 7.
Eight spherical liquid drops join to form a large drop. Diameter of each drop is 2 mm and charge 5µ statC. What is the potential on the surface of the large drop?
Solution:
Radius of a small drop, r = 1 mm 0.1 cm
Suppose, radius of the large drop is R.
∴ \(\frac{4}{3} \pi R^3\) = 8 × \(\frac{4}{3} \pi r^3\) or, R = 2r = 2 × 0.1 = 0.2 cm
∴ Capacitance of large drop, C = R = 0.2 statF
Total charge of the small drops,
Q = 8 × 5 = 40 µstatC = 40 × 10-6 statC
∴ Potential on the surface of the large drop,
V = \(\frac{Q}{C}\) = \(\frac{40 \times 10^{-6}}{0.2}\) statV = 2 × 104 × 300V = 0.06 V
Example 8.
The ratio of the capacitances of two conductors A and B is 2 : 3. The conductor A gains a certain amount of charge and shares It with B. Compare the initial energy of A with the total energy of A and B.
Solution:
Let the capacitance of the conductor A be 2 C and that of the conductor B be 3C.
The amount of charge gained by A is Q.
Let common potential of A and B after sharing of charge be V.
∴ V = \(\frac{Q}{2 C+3 C}\) = \(\frac{Q}{5 C}\)
Energy of the conductor A before sharing of charge,
EA = \(\frac{1}{2} \cdot \frac{Q^2}{2 C}\) = \(\frac{Q^2}{4 C}\)
Total energy of the conductors A and B after sharing the charge,
Example 9.
Each of 27 identical mercury drops is charged to a potential of 10V. If the drops coalesce to form a big drop, what will be its potential? Calculate the ratio of the energy of the big drop to that of a small drop.
Solution:
Let the radius of each small drop be r.
∴ Capacitance, C1 = 4πε0r
∴ Charge of each small drop, q = C1 V1 = 10C1
∴ Total charge Q = 27q = 27 × 10C1 = 270C1
If R be the radius of the big drop, then according to the question,
Example 10.
Charges of 10-2 C and 5 × 10-2 C are put on two metal spheres of radii 1 cm and 2 cm respectively. If they are connected with a metal wire, what will be the final charge on the smaller sphere?
Solution:
Here, radius of first sphere, R1 = 1 cm = 001 m and radius of second sphere, R2 = 2 cm = 0.02 m.
Capacitance of first sphere, C1 = 4πε0R1
and capacitance of second sphere, C2 = 4πε0R2
The total amount of charge betore and after connection,
Q1 + Q2 = Q = (C1 + C2) V [V = common potential]
∴ V = \(\frac{Q_1+Q_2}{C_1+C_2}\)
Final charge on the smaller (first) sphere,
Example 11.
The capacitance and potential, respectively, of a conductor A are 10 unit and 50 unit; those of another conductor B are, respectively, 5 unit and 65 unit. Find out the charges on the two conductors after they are connected by a conducting wire. [HS ‘11]
Solution:
Initially, charge on conductor A ,
Q1 = C1V1 = 10 × 50 = 500 unit
and charge on conductor B,
Q2 = C2V2 = 5 × 65 = 325 unit
The common potential after the two conductors are connected,
V = \(\frac{Q_1+Q_2}{C_1+C_2}\) = \(\frac{500+325}{10+5}\) = \(\frac{825}{15}\) = 55 unit
Now, charge on conductor A,
q1 = C1V = 10 × 55 = 550 unit
and charge on conductor B,
q2 = C2V = 5 × 55 = 275 unit
Example 12.
A spherical liquid drop of capacitance 1µF breaks into 8 drops of the same radius. What is the capacitance of each of these smaller drops? [Karnataka CET ‘04]
Solution:
Let R radius of the initial drop; r = radius of each of 8 smaller drops.
∴ \(\frac{4}{3} \pi R^3\) = \(8 \times \frac{4}{3} \pi r^3\) or, r = \(\frac{R}{2}\)
Capacitance of the bigger drop =4πε0R = 1µF
∴ Capacitance of each small drop
= 4πε0r = 4πε0\(\frac{R}{2}\) = \(\frac{1}{2} \cdot 4 \pi \epsilon_0 F\) = \(\frac{1}{2} \times 1 \mu \mathrm{F}\) = 0.5 µF
Example 13.
Two isolated metallic solid spheres of radii R and 2R are charged in such a way that both of these have same charge density σ. The spheres are placed far away from each other and are connected by a thin conducting wire. Find the new charge density on the bigger sphere.
Solution:
Let σ be the charge density of the two spheres. So, charge of the first sphere = q1 = 4πrR2σ
and charge of the second sphere = q2 = 4π(2R)2r = 16πR2σ
When they are connected with a wire, let \(q_1^{\prime}\) and \(q_2^{\prime}\) be the new charges. Then we may write
\(q_1^{\prime}\) + \(q_2^{\prime}\) = q1 + q2
Since the two spheres are at the same potential,