ROUTERA

Electrostatic Potential and Capacitance

Class 12th Physics Chapter HOTs

1. Question

Question: A quadrupole consists of two dipoles of dipole moment p⃗\vec{p} and −p⃗-\vec{p} separated by a small distance 2a2a . Derive the expression for the electric potential VV at a point on the axis of the quadrupole.

Solution:
The potential at a point due to a dipole is given by:

V=14πϵ0p⃗⋅r^r2V = \frac{1}{4\pi \epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2}

For a quadrupole, the net potential is the superposition of the potentials of the two dipoles:

V=14πϵ0[p(r−a)2−p(r+a)2]V = \frac{1}{4\pi \epsilon_0} \left[ \frac{p}{(r-a)^2} - \frac{p}{(r+a)^2} \right]

Using the binomial approximation for small a/ra/r :

(r±a)2≈r2±2ar(r \pm a)^2 \approx r^2 \pm 2ar V=14πϵ02par3V = \frac{1}{4\pi \epsilon_0} \frac{2p a}{r^3}

Thus,

V=14πϵ0p(2a)r3V = \frac{1}{4\pi \epsilon_0} \frac{p(2a)}{r^3}

2. Question

Question: A parallel plate capacitor has plate area AA and plate separation dd . If a conducting plate of thickness d/2d/2 is inserted symmetrically between the plates, find the new capacitance.

Solution:
The conducting plate divides the capacitor into two capacitors in series, each with plate separation d/2d/2 :

C1=ϵ0Ad/2=2ϵ0AdC_1 = \frac{\epsilon_0 A}{d/2} = \frac{2\epsilon_0 A}{d}

The equivalent capacitance CC of two capacitors in series is:

1C=1C1+1C1=12dϵ0A\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_1} = \frac{1}{2} \frac{d}{\epsilon_0 A} C=ϵ0Ad/2=2ϵ0AdC = \frac{\epsilon_0 A}{d/2} = \frac{2\epsilon_0 A}{d}

3. Question

Question: Derive an expression for the energy density of a parallel plate capacitor in terms of the electric field EE .

Solution:
Energy stored in a capacitor is:

U=12CV2U = \frac{1}{2}CV^2

For a parallel plate capacitor:

C=ϵ0Ad,V=EdC = \frac{\epsilon_0 A}{d}, \quad V = Ed

Substitute CC and VV :

U=12(ϵ0Ad)(Ed)2=12ϵ0AdE2U = \frac{1}{2} \left(\frac{\epsilon_0 A}{d}\right) (Ed)^2 = \frac{1}{2} \epsilon_0 A d E^2

Energy density uu is energy per unit volume:

u=UVolume=12ϵ0AdE2Ad=12ϵ0E2u = \frac{U}{\text{Volume}} = \frac{\frac{1}{2} \epsilon_0 A d E^2}{A d} = \frac{1}{2} \epsilon_0 E^2

4. Question

Question: Why are the equipotential surfaces perpendicular to the electric field lines? Explain with reasoning.

Solution:
Equipotential surfaces are surfaces where the potential is constant. The work done in moving a charge qq along such a surface is zero:

W=qE⃗⋅dr⃗=0W = q \vec{E} \cdot \vec{dr} = 0

This implies that E⃗\vec{E} and dr⃗\vec{dr} are perpendicular, i.e., the electric field is always normal to equipotential surfaces.

5. Question

Question: Calculate the potential energy of a system of three charges q1=+1 μCq_1 = +1\ \mu C , q2=−2 μCq_2 = -2\ \mu C , and q3=+3 μCq_3 = +3\ \mu C placed at the vertices of an equilateral triangle of side a=1 ma = 1\ \text{m} .

Solution:
The potential energy of the system is the sum of potential energies due to each pair:

U=14πϵ0[q1q2r+q2q3r+q3q1r]U = \frac{1}{4\pi \epsilon_0} \left[ \frac{q_1q_2}{r} + \frac{q_2q_3}{r} + \frac{q_3q_1}{r} \right]

Substitute r=a=1 mr = a = 1\ \text{m} and the charge values:

U=9×109[(1)(−2)1+(−2)(3)1+(3)(1)1]U = 9 \times 10^9 \left[ \frac{(1)(-2)}{1} + \frac{(-2)(3)}{1} + \frac{(3)(1)}{1} \right] U=9×109[−2−6+3]=−45×109 JU = 9 \times 10^9 \left[ -2 - 6 + 3 \right] = -45 \times 10^9\ \text{J} U=−45 JU = -45\ \text{J}

6. Question

Question: A parallel plate capacitor has two dielectrics of dielectric constant k1k_1 and k2k_2 , each filling half the area of the capacitor. Derive the expression for the effective capacitance.

Solution:
Each region acts as a separate capacitor in parallel:

C1=ϵ0k1A/2d,C2=ϵ0k2A/2dC_1 = \frac{\epsilon_0 k_1 A/2}{d}, \quad C_2 = \frac{\epsilon_0 k_2 A/2}{d}

The effective capacitance is:

C=C1+C2=ϵ0Ad(k12+k22)C = C_1 + C_2 = \frac{\epsilon_0 A}{d} \left(\frac{k_1}{2} + \frac{k_2}{2}\right) C=ϵ0Adk1+k22C = \frac{\epsilon_0 A}{d} \frac{k_1 + k_2}{2}

7. Question

Question: The potential difference across a 10 cm long conductor is 5 V5\ \text{V} . Calculate the electric field inside the conductor.

Solution:
Electric field EE is the potential gradient:

E=−dVdxE = -\frac{dV}{dx}

Here,

E=50.1=50 V/mE = \frac{5}{0.1} = 50\ \text{V/m}

 

8. Question

Question: A ring of radius RR carries a uniformly distributed charge QQ . Derive the expression for the electric potential at a point on the axis of the ring at a distance xx from its center.

Solution:
At a point on the axis of the ring, all charge elements contribute equally to the potential because of symmetry. The distance of the point from any charge element is:

r=x2+R2r = \sqrt{x^2 + R^2}

The potential due to the ring is:

V=14πϵ0QrV = \frac{1}{4\pi \epsilon_0} \frac{Q}{r}

Substitute rr :

V=14πϵ0Qx2+R2V = \frac{1}{4\pi \epsilon_0} \frac{Q}{\sqrt{x^2 + R^2}}

9. Question

Question: A uniformly charged spherical shell of radius RR carries charge QQ . At what point is the potential maximum, and what is its value?

Solution:
The potential inside a spherical shell is constant and equal to the potential on the surface:

Vsurface=14πϵ0QRV_{\text{surface}} = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}

Outside the shell, the potential decreases as:

V=14πϵ0QrV = \frac{1}{4\pi \epsilon_0} \frac{Q}{r}

Since the potential inside is constant and decreases outside, the maximum value occurs at the surface:

Vmax=14πϵ0QRV_{\text{max}} = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}

10. Question

Question: Derive the expression for the potential VV of an electric dipole at an arbitrary point in space.

Solution:
The potential at a point due to a dipole is:

V=14πϵ0p⃗⋅r^r2V = \frac{1}{4\pi \epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2}

Here,

p⃗⋅r^=pcos⁡θ\vec{p} \cdot \hat{r} = p \cos\theta

where θ\theta is the angle between p⃗\vec{p} and the position vector r^\hat{r} . Thus,

V=14πϵ0pcos⁡θr2V = \frac{1}{4\pi \epsilon_0} \frac{p \cos\theta}{r^2}

11. Question

Question: A 10 μF10\ \mu F capacitor is charged to 100 V100\ \text{V} and then connected in parallel to an uncharged 20 μF20\ \mu F capacitor. Calculate the energy lost during the redistribution of charges.

Solution:
Initial energy:

Uinitial=12C1V2=12(10×10−6)(100)2=0.05 JU_{\text{initial}} = \frac{1}{2} C_1 V^2 = \frac{1}{2} (10 \times 10^{-6})(100)^2 = 0.05\ \text{J}

Final voltage after charge redistribution:

Vf=C1V1C1+C2=10×10010+20=33.33 VV_f = \frac{C_1 V_1}{C_1 + C_2} = \frac{10 \times 100}{10 + 20} = 33.33\ \text{V}

Final energy:

Ufinal=12(C1+C2)Vf2=12(30×10−6)(33.33)2=0.01667 JU_{\text{final}} = \frac{1}{2} (C_1 + C_2) V_f^2 = \frac{1}{2} (30 \times 10^{-6})(33.33)^2 = 0.01667\ \text{J}

Energy lost:

ΔU=Uinitial−Ufinal=0.05−0.01667=0.03333 J\Delta U = U_{\text{initial}} - U_{\text{final}} = 0.05 - 0.01667 = 0.03333\ \text{J}

12. Question

Question: Why does the capacitance of a parallel plate capacitor increase when a dielectric material is introduced, and how does this affect the electric field between the plates?

Solution:
When a dielectric is introduced, the capacitance increases because the dielectric reduces the effective electric field. The reduction occurs due to the polarization of the dielectric, which creates an opposing field. The new capacitance is:

C′=kCC' = kC

where k>1k > 1 is the dielectric constant. The electric field is reduced as:

E′=EkE' = \frac{E}{k}

13. Question

Question: A capacitor has a dielectric of non-uniform permittivity ϵ=ϵ0(1+αx)\epsilon = \epsilon_0 (1 + \alpha x) , where xx is the distance from one plate. Derive the expression for the electric field.

Solution:
From Gauss's law:

∇⃗⋅E⃗=ρfreeϵ\vec{\nabla} \cdot \vec{E} = \frac{\rho_{\text{free}}}{\epsilon}

Substituting ϵ=ϵ0(1+αx)\epsilon = \epsilon_0 (1 + \alpha x) :

dEdx=ρfreeϵ0(1+αx)\frac{dE}{dx} = \frac{\rho_{\text{free}}}{\epsilon_0 (1 + \alpha x)}

Integrating:

E=∫ρfreeϵ0(1+αx)dxE = \int \frac{\rho_{\text{free}}}{\epsilon_0 (1 + \alpha x)} dx E=ρfreeϵ0αln⁡(1+αx)+CE = \frac{\rho_{\text{free}}}{\epsilon_0 \alpha} \ln(1 + \alpha x) + C

14. Question

Question: Derive the expression for the energy stored in a spherical capacitor with inner radius R1R_1 and outer radius R2R_2 .

Solution:
Capacitance of the spherical capacitor:

C=4πϵ0R1R2R2−R1C = \frac{4\pi \epsilon_0 R_1 R_2}{R_2 - R_1}

Energy stored:

U=12CV2U = \frac{1}{2} C V^2

Substitute CC :

U=12(4πϵ0R1R2R2−R1)V2U = \frac{1}{2} \left(\frac{4\pi \epsilon_0 R_1 R_2}{R_2 - R_1}\right) V^2

15. Question

Question: Treat Earth as a spherical conductor of radius 6.4×106 m6.4 \times 10^6\ \text{m} . Calculate its capacitance.

Solution:
The capacitance of a spherical conductor is:

C=4πϵ0RC = 4\pi \epsilon_0 R

Substitute:

C=4π(8.85×10−12)(6.4×106)C = 4\pi (8.85 \times 10^{-12})(6.4 \times 10^6) C=710 μFC = 710\ \mu F

16. Question

Question: Four capacitors C1=4 μFC_1 = 4\ \mu F , C2=6 μFC_2 = 6\ \mu F , C3=8 μFC_3 = 8\ \mu F , and C4=12 μFC_4 = 12\ \mu F are connected as shown in the figure. C1C_1 and C2C_2 are in series, and their combination is in parallel with C3C_3 . This combination is then connected in series with C4C_4 . Find the effective capacitance of the network.

Solution:
Step 1: Combine C1C_1 and C2C_2 in series:

1C12=1C1+1C2=14+16=512\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} C12=125=2.4 μFC_{12} = \frac{12}{5} = 2.4\ \mu F

Step 2: Combine C12C_{12} with C3C_3 in parallel:

C123=C12+C3=2.4+8=10.4 μFC_{123} = C_{12} + C_3 = 2.4 + 8 = 10.4\ \mu F

Step 3: Combine C123C_{123} and C4C_4 in series:

1Ceff=1C123+1C4=110.4+112\frac{1}{C_{\text{eff}}} = \frac{1}{C_{123}} + \frac{1}{C_4} = \frac{1}{10.4} + \frac{1}{12} 1Ceff=0.09615+0.08333=0.17948\frac{1}{C_{\text{eff}}} = 0.09615 + 0.08333 = 0.17948 Ceff=10.17948≈5.57 μFC_{\text{eff}} = \frac{1}{0.17948} \approx 5.57\ \mu F

17. Question

Question: A parallel plate capacitor has plates separated by 5 mm5\ \text{mm} with a potential difference of 100 V100\ \text{V} . A dielectric slab of thickness 3 mm3\ \text{mm} and dielectric constant k=5k = 5 is inserted between the plates. Find the potential difference across the dielectric and the air gap.

Solution:
Electric field in dielectric:

Edielectric=Eairk=1005=20 V/mmE_{\text{dielectric}} = \frac{E_{\text{air}}}{k} = \frac{100}{5} = 20\ \text{V/mm}

Potential difference across the dielectric:

Vdielectric=Edielectric×t=20×3=60 VV_{\text{dielectric}} = E_{\text{dielectric}} \times t = 20 \times 3 = 60\ \text{V}

Potential difference across the air gap:

Vair=V−Vdielectric=100−60=40 VV_{\text{air}} = V - V_{\text{dielectric}} = 100 - 60 = 40\ \text{V}

18. Question

Question: Two identical conducting spheres of radius RR are given charges Q1Q_1 and Q2Q_2 (Q1>Q2Q_1 > Q_2 ). They are then connected by a thin wire. Calculate the final charges on each sphere and the energy lost during redistribution.

Solution:
Final potential is the same for both spheres:

Q1′R=Q2′R\frac{Q_1'}{R} = \frac{Q_2'}{R} Q1′=Q2′=Q1+Q22Q_1' = Q_2' = \frac{Q_1 + Q_2}{2}

Energy before redistribution:

Uinitial=12Q12R+12Q22RU_{\text{initial}} = \frac{1}{2} \frac{Q_1^2}{R} + \frac{1}{2} \frac{Q_2^2}{R}

Energy after redistribution:

Ufinal=2×12(Q1+Q22)2R=(Q1+Q2)24RU_{\text{final}} = 2 \times \frac{1}{2} \frac{\left(\frac{Q_1 + Q_2}{2}\right)^2}{R} = \frac{(Q_1 + Q_2)^2}{4R}

Energy lost:

ΔU=Uinitial−Ufinal\Delta U = U_{\text{initial}} - U_{\text{final}}

Substitute values for Q1,Q2,RQ_1, Q_2, R to calculate.

19. Question

Question: A parallel plate capacitor has plate area 1 m21\ \text{m}^2 and separation 0.01 m0.01\ \text{m} . Air has a dielectric strength of 3×106 V/m3 \times 10^6\ \text{V/m} . What is the maximum charge the capacitor can hold before breakdown?

Solution:
Breakdown voltage:

Vbreakdown=Edielectric×d=3×106×0.01=3×104 VV_{\text{breakdown}} = E_{\text{dielectric}} \times d = 3 \times 10^6 \times 0.01 = 3 \times 10^4\ \text{V}

Capacitance:

C=ϵ0Ad=8.85×10−12×10.01=8.85×10−10 FC = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 1}{0.01} = 8.85 \times 10^{-10}\ \text{F}

Maximum charge:

Q=CV=(8.85×10−10)(3×104)=2.655×10−5 CQ = CV = (8.85 \times 10^{-10})(3 \times 10^4) = 2.655 \times 10^{-5}\ \text{C}

20. Question

Question: Explain how the concept of bound charges inside a dielectric affects the capacitance of a capacitor.

Solution:
When a dielectric is placed in an electric field, its molecules align, creating induced dipoles. This alignment leads to bound charges that oppose the external field, effectively reducing the field inside the dielectric. As a result, for the same applied voltage, more charge is stored on the capacitor plates. Hence, the capacitance increases by a factor of the dielectric constant kk :

C′=kCC' = kC

This is why dielectric materials are used to enhance the energy storage of capacitors.

21. Question

Question: Derive the expression for the capacitance of a single isolated spherical conductor of radius RR .
Also, calculate the capacitance if R=10 cmR = 10\ \text{cm} .

Solution:
For a spherical conductor, the electric potential at its surface is:

V=Q4πϵ0RV = \frac{Q}{4\pi \epsilon_0 R}

Capacitance is defined as:

C=QV=4πϵ0R1C = \frac{Q}{V} = \frac{4\pi \epsilon_0 R}{1}

Thus, the capacitance of an isolated spherical conductor is:

C=4πϵ0RC = 4\pi \epsilon_0 R

Substitute ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12}\ \text{F/m} and R=0.1 mR = 0.1\ \text{m} :

C=4π(8.85×10−12)(0.1)=1.11×10−11 F=11.1 pFC = 4\pi (8.85 \times 10^{-12}) (0.1) = 1.11 \times 10^{-11}\ \text{F} = 11.1\ \text{pF}

22. Question

Question: A parallel plate capacitor of plate area A=200 cm2A = 200\ \text{cm}^2 and separation d=0.5 cmd = 0.5\ \text{cm} is connected to a 500 V500\ \text{V} battery. A dielectric slab with k=4k = 4 and thickness equal to the plate separation is inserted between the plates. Calculate the energy stored in the capacitor before and after inserting the dielectric.

Solution:
Without the dielectric:

C=ϵ0Ad=8.85×10−12×200×10−40.005=3.54×10−12 FC = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 200 \times 10^{-4}}{0.005} = 3.54 \times 10^{-12}\ \text{F}

Energy stored:

U=12CV2=12(3.54×10−12)(500)2=4.425×10−7 JU = \frac{1}{2} C V^2 = \frac{1}{2} (3.54 \times 10^{-12})(500)^2 = 4.425 \times 10^{-7}\ \text{J}

With the dielectric:

C′=kC=4×3.54×10−12=1.416×10−11 FC' = kC = 4 \times 3.54 \times 10^{-12} = 1.416 \times 10^{-11}\ \text{F} U′=12C′V2=12(1.416×10−11)(500)2=1.77×10−6 JU' = \frac{1}{2} C' V^2 = \frac{1}{2} (1.416 \times 10^{-11})(500)^2 = 1.77 \times 10^{-6}\ \text{J}

23. Question

Question: A capacitor is charged to V=200 VV = 200\ \text{V} and then disconnected from the battery. A dielectric of k=3k = 3 is inserted between its plates. Calculate the energy lost during this process.

Solution:
Initial capacitance:

C=QVC = \frac{Q}{V}

Initial energy:

Uinitial=12CV2U_{\text{initial}} = \frac{1}{2} C V^2

After inserting the dielectric:

C′=kCC' = kC

Charge remains constant, so:

Ufinal=Q22C′=12CV2kU_{\text{final}} = \frac{Q^2}{2C'} = \frac{1}{2} \frac{C V^2}{k}

Energy lost:

ΔU=Uinitial−Ufinal=12CV2(1−1k)\Delta U = U_{\text{initial}} - U_{\text{final}} = \frac{1}{2} C V^2 \left(1 - \frac{1}{k}\right)

Substitute values to calculate.

24. Question

Question: Derive the expression for the energy density of an electric field in a parallel plate capacitor.

Solution:
Energy stored in a capacitor:

U=12CV2U = \frac{1}{2} C V^2

Capacitance of a parallel plate capacitor:

C=ϵ0AdC = \frac{\epsilon_0 A}{d} U=12ϵ0AdV2U = \frac{1}{2} \frac{\epsilon_0 A}{d} V^2

Electric field:

E=Vd  ⟹  V2=E2d2E = \frac{V}{d} \implies V^2 = E^2 d^2

Substitute V2V^2 :

U=12ϵ0AdE2U = \frac{1}{2} \epsilon_0 A d E^2

Energy density:

u=UVolume=12ϵ0AdE2Ad=12ϵ0E2u = \frac{U}{\text{Volume}} = \frac{\frac{1}{2} \epsilon_0 A d E^2}{A d} = \frac{1}{2} \epsilon_0 E^2

25. Question

Question: A capacitor of 10 μF10\ \mu F is charged through a 100 Ω100\ \Omega resistor using a 12 V12\ \text{V} battery. Calculate the voltage across the capacitor after t=1 mst = 1\ \text{ms} .

Solution:
Voltage across the capacitor in an RC circuit:

V(t)=V0(1−e−tRC)V(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right) R=100 Ω, C=10×10−6 F, V0=12 VR = 100\ \Omega,\ C = 10 \times 10^{-6}\ \text{F},\ V_0 = 12\ \text{V}

Time constant:

τ=RC=(100)(10×10−6)=10−3 s\tau = RC = (100)(10 \times 10^{-6}) = 10^{-3}\ \text{s}

Substitute t=1 ms=10−3 st = 1\ \text{ms} = 10^{-3}\ \text{s} :

V(t)=12(1−e−10−310−3)=12(1−e−1)V(t) = 12 \left(1 - e^{-\frac{10^{-3}}{10^{-3}}}\right) = 12 \left(1 - e^{-1}\right) e−1≈0.3679e^{-1} \approx 0.3679 V(t)=12×(1−0.3679)=12×0.6321=7.58 VV(t) = 12 \times (1 - 0.3679) = 12 \times 0.6321 = 7.58\ \text{V}

26. Question

Question: Why does a cavity inside a conductor have no electric field, irrespective of the external electric field?

Solution:
The free charges in a conductor redistribute themselves in response to any external electric field. This redistribution creates an induced electric field inside the conductor, exactly canceling the external field. Hence, the net electric field inside the conductor, and any cavity within it, is zero. This phenomenon is called electrostatic shielding.

27. Question

Question: Find the equivalent capacitance between points AA and BB in the following arrangement:
Three capacitors, C1=4 μFC_1 = 4\ \mu\text{F} , C2=6 μFC_2 = 6\ \mu\text{F} , and C3=8 μFC_3 = 8\ \mu\text{F} , are connected in such a way that C1C_1 and C2C_2 are in series, and their combination is in parallel with C3C_3 .

Solution:

  1. Capacitance of C1C_1 and C2C_2 in series:
1Cseries=1C1+1C2=14+16=312+212=512\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} Cseries=125=2.4 μFC_{\text{series}} = \frac{12}{5} = 2.4\ \mu\text{F}
  1. Capacitance of CseriesC_{\text{series}} and C3C_3 in parallel:
Ceq=Cseries+C3=2.4+8=10.4 μFC_{\text{eq}} = C_{\text{series}} + C_3 = 2.4 + 8 = 10.4\ \mu\text{F}

The equivalent capacitance is:

Ceq=10.4 μFC_{\text{eq}} = 10.4\ \mu\text{F}

28. Question

Question: A 2 μF2\ \mu\text{F} capacitor charged to 500 V500\ \text{V} is discharged through a resistor of 50 Ω50\ \Omega . Calculate the energy dissipated as heat in the resistor.

Solution:
The total energy stored in the capacitor:

U=12CV2U = \frac{1}{2} C V^2

Substitute C=2×10−6 FC = 2 \times 10^{-6}\ \text{F} and V=500 VV = 500\ \text{V} :

U=12(2×10−6)(500)2=12(2×10−6)(250000)=0.25 JU = \frac{1}{2} (2 \times 10^{-6}) (500)^2 = \frac{1}{2} (2 \times 10^{-6}) (250000) = 0.25\ \text{J}

Since all the energy stored in the capacitor is dissipated as heat in the resistor, the energy dissipated is:

0.25 J0.25\ \text{J}

29. Question

Question: Two capacitors, C1=6 μFC_1 = 6\ \mu\text{F} and C2=12 μFC_2 = 12\ \mu\text{F} , are connected in series to a 120 V120\ \text{V} battery. Calculate the total energy stored in the system.

Solution:

  1. Equivalent Capacitance of the Combination:
1Ceq=1C1+1C2=16+112=212+112=312\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} Ceq=123=4 μFC_{\text{eq}} = \frac{12}{3} = 4\ \mu\text{F}
  1. Energy Stored:
U=12CeqV2U = \frac{1}{2} C_{\text{eq}} V^2

Substitute Ceq=4×10−6 FC_{\text{eq}} = 4 \times 10^{-6}\ \text{F} and V=120 VV = 120\ \text{V} :

U=12(4×10−6)(120)2=12(4×10−6)(14400)=0.0288 JU = \frac{1}{2} (4 \times 10^{-6}) (120)^2 = \frac{1}{2} (4 \times 10^{-6}) (14400) = 0.0288\ \text{J}

The total energy stored is:

0.0288 J0.0288\ \text{J}

30. Question

Question: A parallel plate capacitor is connected to a battery and fully charged. If a conductor of negligible thickness is introduced midway between the plates, how will the capacitance, charge, and energy stored change?

Solution:

  1. Effect on Capacitance:
    Introducing a conducting plate divides the capacitor into two capacitors in series, each with half the plate separation (d/2d/2 ).
    The capacitance of each is:
C′=ϵ0Ad/2=2ϵ0Ad=2CC' = \frac{\epsilon_0 A}{d/2} = 2 \frac{\epsilon_0 A}{d} = 2C

For two capacitors in series:

1Ceq=1C′+1C′=12C+12C=1C\frac{1}{C_{\text{eq}}} = \frac{1}{C'} + \frac{1}{C'} = \frac{1}{2C} + \frac{1}{2C} = \frac{1}{C} Ceq=CC_{\text{eq}} = C

Thus, the capacitance does not change.

  1. Effect on Charge:
    Since the capacitance remains unchanged and the voltage remains constant, the charge Q=CVQ = CV also remains unchanged.

  2. Effect on Energy:
    Energy stored:

U=12CV2U = \frac{1}{2} CV^2

Since CC and VV are constant, the energy stored also remains unchanged.

The system's capacitance, charge, and energy remain the same, but the internal electric field distribution may change due to the presence of the conductor.