108 + Interview Questions in CAPACITORS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Find the capacitance of a capacitor having a charge of `6xx10^(-7)C` and energy of `4.5xx10^(-4)J`
Answer» Correct Answer - 400pF

Discussion

2.	Calculate the capacitance per metre of a capacitor formed by two long coaxial cylinders of radii 5 and 5.2 cm filled with dielectric of permittivity 1.2.
Answer» Correct Answer - `1.7xx10^(-9)F`

Discussion

3.	A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10cm of redii 2mm and for mm. (a) calculate the capacitance (b) another capacitor of the same length is constructed with cylinders of radii 4mm and 8mm. Calculate the capacitance .
Answer» Correct Answer - 8pF

Discussion

4.	Find the capacitance of unit length of a cylindrical capacitor which has conductor of radii 2.5 mm and 4.5 mm, respectively and the dielectric consists of two layers whose cylinder of contact is 3.5 cm in radius, the inner layer having a dielectric coefficinet 4 and the outer layer 6.
Answer» Correct Answer - `0.442xx10^(-9)`F

Discussion

5.	If the capacitance of a conductor carrying a charge of 8 C is 0.005 F, calculate its potential.
Answer» Correct Answer - 1600V

Discussion

6.	If 64 drps each charged to 220 V coalesce, what will be the potential of the bigger drop?
Answer» Correct Answer - 3520V

Discussion

7.	A parallel-plate capacitor having plate are `100 cm ^2` and separation 1.0 mm holds a charge of `0.12 muC` when connected to a 120 V battery. Find the dielectric constant of the material filling the gap.
Answer» Correct Answer - 11.3

Discussion

8.	A capacitor charged from a 50 V d.c. supply is found to have charge of `10 muC`. What is the capacitance of the capacitor and how much energy is stored in it?
Answer» Correct Answer - 0.2 `muF,2.5xx10^(-4)J`

Discussion

9.	Capacitance of a conductor is `1muF`. What charge is required to raise its potential to `100 V`?
Answer» Correct Answer - A::C Using the equation `q=CV` we have, `q=(1muF)(100V)` `=100muC`

Discussion

10.	A sphere of radius `0.03m` is suspended within a hollow sphere of radius `0.05m`. If the inner sphere is charged to a potential of `1500` volt and outer sphere is earthed. Find the capacitance and the charge of the inner sphere.
Answer» Correct Answer - `8.33pF,1.25xx10^(-16)C`

Discussion

11.	Find the capacitnace of a conducting sphere of radius 10 cm situated in air. How much charge is required to raise it to a potential 1000 volt?
Answer» Correct Answer - `11pF,1.1xx10^(-8)C`

Discussion

12.	A capacitor having a capacitance of 100μF is charged to a potential difference of 50V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Calculate the new potential difference between the plates, (c) What charge would have produced this potential .difference in absence of the dielectric slab, (d) Find the charge induced at a surface of the dielectric slab.
Answer» Capacitance = 100μF = 10^–4F P.d = 30V (a) q = CV = 10^–4 × 50 = 5 × 10^–3 c = 5mc Dielectric constant = 2.5 (b) New C = C' = 2.5 × C = 2.5 × 10^–4F New p.d = q/c₁ [∴’q’ remains same after disconnection of battery] = (5 x 10^-3)/(2.5 x 10^-4) = 20V. (c) In the absence of the dielectric slab, the charge that must have produced C × V = 10^–4 × 20 = 2 × 10^–3 c = 2mc (d) Charge induced at a surface of the dielectric slab = q(1 –1/k) (where k = dielectric constant, q = charge of plate) = 5 × 10^–3(1 - 1/2.5) = 5 x 10^-3 x 3/5 = 3 x 10^-3 = 3mc.

12.

A capacitor having a capacitance of 100μF is charged to a potential difference of 50V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Calculate the new potential difference between the plates, (c) What charge would have produced this potential .difference in absence of the dielectric slab, (d) Find the charge induced at a surface of the dielectric slab.

Answer»

Capacitance = 100μF = 10^–4F

P.d = 30V

(a) q = CV = 10^–4 × 50 = 5 × 10^–3 c = 5mc

Dielectric constant = 2.5

(b) New C = C' = 2.5 × C = 2.5 × 10^–4F

New p.d = q/c₁ [∴’q’ remains same after disconnection of battery]

= (5 x 10^-3)/(2.5 x 10^-4) = 20V.

(c) In the absence of the dielectric slab, the charge that must have produced

C × V = 10^–4 × 20 = 2 × 10^–3 c = 2mc

(d) Charge induced at a surface of the dielectric slab

= q(1 –1/k) (where k = dielectric constant, q = charge of plate)

= 5 × 10^–3(1 - 1/2.5) = 5 x 10^-3 x 3/5 = 3 x 10^-3 = 3mc.

Discussion

13.	Two capacitor each having capacitance C and breakdown voltage V are joined in series. the capacitance and the breakdown voltage of the combination will be (a) 2C and 2V (b) C/2 and V/2(c) 2C and V/2(d) C/2 and V.
Answer» The correct answer is (d) C/2 and V.

Discussion

14.	A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is(a) CV/ε0(b) 2CV/ε0(c) CV/2ε0(d) zero.
Answer» The correct answer is (d) zero.

Discussion

15.	Three capacitors of 10,15 and 30 `muF` are connected in series and on this combination a potential difference of 60 V is applied. Calculate the charge, potential difference and energy stored on each capacitor.
Answer» Correct Answer - `q_(1)=q_(2)=q_(3)=3xx10^(-4)C,V_(1)=30V,V_(2)=20V,V_(3)=10V, U_(1)=4.5xx10^(-3)J,U_(2)=3.0xx10^(-3)J,U_(3)=1.5xx10^(-3)J`

Discussion

16.	Assertion : In series combination, charges on two capacitors are always equal. Reason : If charges are same, the total potential difference applied across two capacitors will be distributed in inverse ratio of capacities.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true.
Answer» Correct Answer - D Charges are same, if initially the capacitors are uncharged. Further, `V=q/C` Hence `Vprop1/C` if `q` same

Discussion

17.	find the charges on the capacitors in figure. And the potential differences across them.
Answer» Correct Answer - `q_(1)=144xx10^(-6)C,q_(2)=96xx10^(-6)C,q_(3)=48xx10^(-6)C,V_(xy)=72V,V_(yz)=48V`

Discussion

18.	`n` identical capacitors are connected in parallel to a potential difference `V`. These capacitors are then reconnected in sereis, their charges being left undisturbed. The potential difference obtained isA. zeroB. `(n-1)V`C. `nV`D. `n^2V`
Answer» Correct Answer - C `V_(n et)=V_1+V_2+……….` (in series) `=V+V+…….` `nV`

Discussion

19.	The electric field on two sides of a thin sheet of charge is shown in the figure. The charge density on the sheet is A. `2epsilon_0`B. `4epsilon_0`C. `10epsilon_0`D. zero
Answer» Correct Answer - B Let `E` be the external field (toward right). The `E-sigma/(2epsilon_0) =8`………….i `E+sigma/(2epsilon_0=12`………..ii Solving these equations we get `sigma=4epsilon_0`

Discussion

20.	The energy density in the electric field created by a point charge falls off with the distance from the point charge as (a) 1/r (b) 1/r2(c) 1/r3(d) 1/r4.
Answer» The correct answer is (d) 1/r⁴.

Discussion

21.	A parallel-plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let Q, and Q be the charges appearing on the positive and negative plates respectively, (a) Q > Q (b)Q = Q (c) Q < Q(d) The information is not sufficient to decide the relation between Q and Q.
Answer» The correct answer is (b)Q = Q

Discussion

22.	A thin metal plate P is inserted between the plates of a parallel plate capacitor of capacitance C in such a way that its edges touch the two plates (figure 31-Q2). The capacitance now becomes(a) C/2 (b) 2C (c) 0 (d) ∞
Answer» The correct answer is (d) ∞

Discussion

23.	Three capacitors of capacitances 6μF each are available. The minimum and maximum capacitances, which m&y be obtained are (a) 6μF, 18μF (b) 3μF, 12μF (c) 2μF, 12μF (d) 2μF, 18μF.
Answer» The correct answer is (d) 2μF, 18μF.

Discussion

24.	Figure (31-Q3) shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors. (a) C1 > C2. (b) C1 = C2.(c) C1 < C2. (d) The information is not sufficient to decide the relation between C1 and C2.
Answer» The correct answer is (c) C₁ < C₂.

Discussion

25.	Two metal spheres of capacitances C1 and C2, carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy(a) Q1/Q2 < C1/C2 (b) Q1/Q2 = C1/C2(c) Q1/Q2 > C1/c2(d) Q1/Q2 = C2/C1.
Answer» The correct answer is (b) Q₁/Q₂ = C₁/C₂

Discussion

26.	Two metal plates having charges Q, - Q face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will (a) increase (b) decrease (c) remain the same (d) become zero.
Answer» (a) increase

Discussion

27.	Assertion : When an uncharged capacitor is charged by a battery only `50%` of the energy supplied is stored in the capacitor. Reason : Rest `50%` is lost.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true.
Answer» Correct Answer - A::B Energy supplied by the battery is `/_qV=(CV)(V)=CV^2` Energy stored in the capacitor is `1/2CV^2`

Discussion

28.	How many time constants will elapse before the current in a charging `R-C` circuit drops to half of its initial value?
Answer» `i=i_0e^(-t/tau_C)` putting `i=i_0/2`, we get `t=(ln 2)tau_C=(0.693)tau_C`

Discussion

29.	An Uncharged capacitor `C` is connected to a battery through a resistance `R`. Show that by the time the capacitor gets fully charged, the energy dissipated in ``R is the same as the energy stored in `C`.
Answer» Correct Answer - A Charge supplied by the battery `q=CV` Energy supplied by the battery `E=qV=CV^2` Energy stored in the capacitor `U=1/2CV^2` `:.` Energy dissipated across `R` in the form of heat `=E-U=1/2CV^2=u`

Discussion

30.	A charge of 15 nC (nanocoulomb) raises the potential of a spherical from 500 V to 1500 V Find the radius of the conductor.
Answer» Correct Answer - `13.5 cm`

Discussion

31.	A parallel-plate condenser of area 100 `cm^(2)`, plate separation 5 mm is charged to a p.d. of 100 V when air is used between the plates. If now air is replaced by glass `(epsilon_(r)=6)` calculate (a) the new capacity, (b) the new p.d. and (c) the loss of energy. Account for the loss of energy.
Answer» Correct Answer - `(a)1.062xx10^(-10)F,(b)16.7V,(c)7.37xx10^(-8)J`

Discussion

32.	A graph between current and time during charging of a capacitor by a battery in series with a resistor is shown. The graphs are drawn for two circutis. `R_1, R_2, C_1, C_2` and `V_1, V_2` are the values of resistance, capacitance and `EMF` of the cell in the two circuits. If only two parameters (out of resistance, capacitance, `EMF`) are different in the two circuits. What may be the correct options (s) ? A. `V_1=V_2,R_1gtR_2,C_1gtC_2`B. `V_1gtV_2,R_1gtR_2,C_1=C_2`C. `V_1ltV_2,R_1ltR_2,C_1=C_2`D. `V_1ltV_2,R_1=R_2,C_1,C_2`
Answer» Correct Answer - C Changing current is given by `i=i_0e^(-t//tau_C)` or `i=V/Re^(-t//CR)` If we have take log on both sides, we have `ln(i)=ln(V/R)-(1/(CR))t` Hence ln(i) versus `t` graph is a straight line with slope `(-1/(CR))` and intercept `+ln(V/R)` . Intercepts are same, but `\|slope_i\|gt\|slope\|_2`.

Discussion

33.	The figure shows a graph of the current in a charging circuit of a capacitor through a resistor of resistance `10Omega`. A. the initial potential difference across the capacitor is `100V`B. The capacitor of the capcitor is `1/(10ln2) F`C. the total heat produced in the circuit will be `(500/ln2)J`D. All of the above
Answer» Correct Answer - D `V_0=i_0R=(10)(10)=100V` after `2s` , current becoms `1/4` th. Therefore, after `1s`, current will remain half also called half life. `t_(1/2)=(ln2)tau_C=(ln 2)CR` `:. C=((t_(1/2)))/((ln 2)R)=1/(10 ln 2) F` total heat `=1/2 CV_0^2` `=1/2x1/(10 ln 2) (100)^2` `=500/(ln 2) J`

Discussion

34.	A capacitor of capacity `C` is charged to a potential difference `V` and another capacitor of capacity `2C` is charged to a potential difference `4V`. The charged batteries are disconnected and the two capacitors are connected with reverse polarity (i.e. positive plate of first capacitor is connected to negative plate of second capacitor). The heat produced during the redistribution of charge between the capacitors will beA. `(125CV^2)/3`B. `(50CV^2)/3`C. `2CV^2`D. `(25CV^2)/3`
Answer» Correct Answer - D Total charge=`(2C)(4V)-CV` `=7CV` `V_C=("Total charge")/("Total capacitance")` `(7CV)/(2C+C)=7/3V` Heat `=U_i+U_f` `=1/2CV^2+1/2(2C)(4V)^2-1/2xx3Cxx(7/3V)^2` `=25/3CV^2`

Discussion

35.	Four capacitors are connected inseries a battery of emf `10 V` as shown in the figure. The point `P` is earthed. The potential of point `A` is equal in magnitude to potetial of points `B` but opposite in sign is A. `C_1+C_2+C_3=C_4`B. `1/C_1+1/C_2+1/C_3=1/C_4`C. `(C_1C_2C_3)/(C_1^2+C_2^2+C_3^2)=C_4`D. its is never possible
Answer» Correct Answer - B Net capacitance between points `A` and `P` will be equal to the net capacitance between points `P` and `B`.

Discussion

36.	Two metal balls with radii `r_(1)=1 cm` and `r_(2)=2cm` at a large distance R=100 cm from each other, are connected to a battery of emf `epsi=3000 V` Find the of interactio betwenn the balls. What is the signification of the ward large?
Answer» Correct Answer - `epsi^(2)((r_(1)r_(2))/(r_(1)+r_(2)))^(2)/(R^(2))=4.4xx10^(-8)N;` since the distance is large, charge is assumed to be unifromly distribuited

Discussion

37.	Two long straight wires with equal cross-sectional radii a area located parallel to each other in air at large distance b from each other. Find the capacitance per unit length of the wires. [Hint: Take charge to be uniformly distributed. Consider the field at any point and hence find the p.d. between the wires.]
Answer» Correct Answer - `(piepsilon_(0))/(In b//d)`

Discussion

38.	A capacitor having a capacitance of 100μF is charged to a potential difference of 24V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12V with the positive plate of the capacitor joined with the positive terminal of the battery, (a) Find the charges on the capacitor before and after the reconnection. (b) Find the charge flown through the 12V battery, (c) Is work done by the battery or is it done on the battery? Find its magnitude, (d) Find the decrease in electrostatic field energy, (e) Find the heat developed during the flow of charge after reconnection.
Answer» (a) Before reconnection C = 100μf V = 24V q = CV = 2400μc (Before reconnection) After connection When C = 100μf V = 12V q = CV = 1200μc (After connection) (b) C = 100, V = 12V ∴ q = CV = 1200v (c) We know V = W/q W = vq = 12 × 1200 = 14400J = 14.4mJ The work done on the battery. (d) Initial electrostatic field energy Ui = (1/2) CV₁² Final Electrostatic field energy Uf = (1/2) CV₂² ∴ Decrease in Electrostatic Field energy = (1/2) CV₁²– (1/2) CV₂² = (1/2)C(V₁² – V₂²) = (1/2) × 100(576 –144) = 21600J ∴ Energy = 21600j = 21.6mJ (e)After reconnection C = 100μc, V = 12v ∴ The energy appeared = (1/2)CV₂ = (1/2) × 100 × 144 = 7200J = 7.2mJ This amount of energy is developed as heat when the charge flow through the capacitor.

38.

A capacitor having a capacitance of 100μF is charged to a potential difference of 24V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12V with the positive plate of the capacitor joined with the positive terminal of the battery, (a) Find the charges on the capacitor before and after the reconnection. (b) Find the charge flown through the 12V battery, (c) Is work done by the battery or is it done on the battery? Find its magnitude, (d) Find the decrease in electrostatic field energy, (e) Find the heat developed during the flow of charge after reconnection.

Answer»

(a) Before reconnection

C = 100μf

V = 24V

q = CV = 2400μc (Before reconnection)

After connection

When C = 100μf

V = 12V

q = CV = 1200μc (After connection)

(b) C = 100, V = 12V

∴ q = CV = 1200v

(c) We know V = W/q

W = vq = 12 × 1200 = 14400J = 14.4mJ The work done on the battery.

(d) Initial electrostatic field energy Ui = (1/2) CV₁²

Final Electrostatic field energy Uf = (1/2) CV₂²

∴ Decrease in Electrostatic

Field energy = (1/2) CV₁²– (1/2) CV₂²

= (1/2)C(V₁² – V₂²) = (1/2) × 100(576 –144) = 21600J

∴ Energy = 21600j = 21.6mJ

(e)After reconnection

C = 100μc,

V = 12v

∴ The energy appeared = (1/2)CV₂ = (1/2) × 100 × 144 = 7200J = 7.2mJ

This amount of energy is developed as heat when the charge flow through the capacitor.

Discussion

39.	Calculate the self potential energy of charge q distributed over the surface of a hollow sphere of radius R. What happens to this self energy when R=0, or when the charge is an ideal point charge?
Answer» Correct Answer - `(q^(2))/(8piepsilon_(0)R)`

Discussion

40.	A charge q is distributed over the volume of a sphere of radius R. Assuming the permittivty to be equal to unity, compare the energy stored inside the sphere with the energy pervading the surrounding space. [Hint: `u=(1)/(2)epsilon_(0)E^(2).`]
Answer» Correct Answer - `1:5`

Discussion

41.	Four capacitors of equal cpacitances are connected in series with a battery of 10 V, as shown in figure. The middle point B is connected to the earth. What will be the potentials of the points A and C?
Answer» Correct Answer - `V_(A)=+5V,V_(C)=-5V`

Discussion

42.	A capacitors `C_1` is charged to a potential `V` and connected to another capacitor in seris with a resistor `R` as shown. It is observed that heat `H_1` is dissipated across resistance `R`, tilll the circuit reaches steady state. Same process is repeated using resistance of `2R`. If `H_2` is heat dissipated in this case then A. `H_2/H_1=1`B. `H_2/H_1=4`C. `H_2/H_1=1/4`D. `H_2/H_1=2`
Answer» Correct Answer - A `H_1=H_2=U_i-U_f` the only change is by increasing the resistance `tau_C` increase. Hence process of radistribution of charge slows down.

Discussion

43.	Find the dimensions of capacitance.
Answer» `U=1/2q^2/C` `:. [C]=[q^2/U]=[(A^2T^2)/(ML^2T^-2)]` `=[M^-1L^-2T^4A^2]`

Discussion

44.	No charge will flow when two conductors having the same charge are connected to each other. this statement true or false?
Answer» Charge does not flow if their potentials are same.

Discussion

45.	A parallel plate capacitor has capacitance of `1.0 F`. If the plates are `1.0 mm` apart, what is the area of the plates?
Answer» Correct Answer - `113m^(2)`

Discussion

46.	A capacitor has a capacitance of `7.28 muF`. What amount of charge must be placed on each of its plates to make the potential difference between its plates equAl to `25.0 V`?
Answer» Correct Answer - `q=CV`

Discussion

47.	Find the charges on different capacitors.
Answer» `C_(n et)=3muF, q=CV=3xx10=120muC` Now, this `q` will be distributed between `9muF` and `3muF` in the direct ratio of their capcities.

Discussion

48.	The potential difference between points `a` and `b` of circuits shown in figure is A. `((E_1+E_2)/(C_1+C_2))C_2`B. `((E_2-E_2)/(C_1+C_2))C_2`C. `((E_1+E_2)/(C_1+C_2))C_1`D. `((E_1-E_2)/(C_1+C_2))C_1`
Answer» Correct Answer - C `q=(E_1+E_2)C_(n et)` `=(E_1+E_2)(C_1C_2)/(C_2+C_2)` `V_(ab)=q/C_2=((E_1+E_2)/(C_1+C_2))C_1`

Discussion

49.	A parallel air capacitance `245muF` has a charge of magnitude `0.48muC` on each plate. The plates are `0.328 mm` apart. (a) What is the potential difference between the plates? (b) What is the area of each plate?(c) What is the surface charge density on each plate?
Answer» `a. V=q/C` b. `C=(epsilon_0A)/d` `:. A=(Cd)/epsilon_0` `c. sigma=q/A`

Discussion

50.	Find charges on different capacitors.
Answer» `C_(n et)=2muF, q=CV=2xx15=30muC` now this `q` will be distributed between `4muF` and `2muF` in direct ratio of their capacities.

Discussion

Explore topic-wise InterviewSolutions in .

Find the capacitance of a capacitor having a charge of `6xx10^(-7)C` and energy of `4.5xx10^(-4)J`

Calculate the capacitance per metre of a capacitor formed by two long coaxial cylinders of radii 5 and 5.2 cm filled with dielectric of permittivity 1.2.

A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10cm of redii 2mm and for mm. (a) calculate the capacitance (b) another capacitor of the same length is constructed with cylinders of radii 4mm and 8mm. Calculate the capacitance .

Find the capacitance of unit length of a cylindrical capacitor which has conductor of radii 2.5 mm and 4.5 mm, respectively and the dielectric consists of two layers whose cylinder of contact is 3.5 cm in radius, the inner layer having a dielectric coefficinet 4 and the outer layer 6.

If the capacitance of a conductor carrying a charge of 8 C is 0.005 F, calculate its potential.

If 64 drps each charged to 220 V coalesce, what will be the potential of the bigger drop?

A parallel-plate capacitor having plate are `100 cm ^2` and separation 1.0 mm holds a charge of `0.12 muC` when connected to a 120 V battery. Find the dielectric constant of the material filling the gap.

A capacitor charged from a 50 V d.c. supply is found to have charge of `10 muC`. What is the capacitance of the capacitor and how much energy is stored in it?

Capacitance of a conductor is `1muF`. What charge is required to raise its potential to `100 V`?

A sphere of radius `0.03m` is suspended within a hollow sphere of radius `0.05m`. If the inner sphere is charged to a potential of `1500` volt and outer sphere is earthed. Find the capacitance and the charge of the inner sphere.

Find the capacitnace of a conducting sphere of radius 10 cm situated in air. How much charge is required to raise it to a potential 1000 volt?

Two capacitor each having capacitance C and breakdown voltage V are joined in series. the capacitance and the breakdown voltage of the combination will be (a) 2C and 2V (b) C/2 and V/2(c) 2C and V/2(d) C/2 and V.

A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is(a) CV/ε0(b) 2CV/ε0(c) CV/2ε0(d) zero.

Three capacitors of 10,15 and 30 `muF` are connected in series and on this combination a potential difference of 60 V is applied. Calculate the charge, potential difference and energy stored on each capacitor.

find the charges on the capacitors in figure. And the potential differences across them.

`n` identical capacitors are connected in parallel to a potential difference `V`. These capacitors are then reconnected in sereis, their charges being left undisturbed. The potential difference obtained isA. zeroB. `(n-1)V`C. `nV`D. `n^2V`

The electric field on two sides of a thin sheet of charge is shown in the figure. The charge density on the sheet is A. `2epsilon_0`B. `4epsilon_0`C. `10epsilon_0`D. zero

The energy density in the electric field created by a point charge falls off with the distance from the point charge as (a) 1/r (b) 1/r2(c) 1/r3(d) 1/r4.

A thin metal plate P is inserted between the plates of a parallel plate capacitor of capacitance C in such a way that its edges touch the two plates (figure 31-Q2). The capacitance now becomes(a) C/2 (b) 2C (c) 0 (d) ∞

Three capacitors of capacitances 6μF each are available. The minimum and maximum capacitances, which m&amp;y be obtained are (a) 6μF, 18μF (b) 3μF, 12μF (c) 2μF, 12μF (d) 2μF, 18μF.

Two metal spheres of capacitances C1 and C2, carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy(a) Q1/Q2 &lt; C1/C2 (b) Q1/Q2 = C1/C2(c) Q1/Q2 &gt; C1/c2(d) Q1/Q2 = C2/C1.

Two metal plates having charges Q, - Q face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will (a) increase (b) decrease (c) remain the same (d) become zero.

How many time constants will elapse before the current in a charging `R-C` circuit drops to half of its initial value?

An Uncharged capacitor `C` is connected to a battery through a resistance `R`. Show that by the time the capacitor gets fully charged, the energy dissipated in ``R is the same as the energy stored in `C`.

A charge of 15 nC (nanocoulomb) raises the potential of a spherical from 500 V to 1500 V Find the radius of the conductor.

A parallel-plate condenser of area 100 `cm^(2)`, plate separation 5 mm is charged to a p.d. of 100 V when air is used between the plates. If now air is replaced by glass `(epsilon_(r)=6)` calculate (a) the new capacity, (b) the new p.d. and (c) the loss of energy. Account for the loss of energy.

Two metal balls with radii `r_(1)=1 cm` and `r_(2)=2cm` at a large distance R=100 cm from each other, are connected to a battery of emf `epsi=3000 V` Find the of interactio betwenn the balls. What is the signification of the ward large?

Calculate the self potential energy of charge q distributed over the surface of a hollow sphere of radius R. What happens to this self energy when R=0, or when the charge is an ideal point charge?

A charge q is distributed over the volume of a sphere of radius R. Assuming the permittivty to be equal to unity, compare the energy stored inside the sphere with the energy pervading the surrounding space. [Hint: `u=(1)/(2)epsilon_(0)E^(2).`]

Four capacitors of equal cpacitances are connected in series with a battery of 10 V, as shown in figure. The middle point B is connected to the earth. What will be the potentials of the points A and C?

Find the dimensions of capacitance.

No charge will flow when two conductors having the same charge are connected to each other. this statement true or false?

A parallel plate capacitor has capacitance of `1.0 F`. If the plates are `1.0 mm` apart, what is the area of the plates?

A capacitor has a capacitance of `7.28 muF`. What amount of charge must be placed on each of its plates to make the potential difference between its plates equAl to `25.0 V`?

Find the charges on different capacitors.

The potential difference between points `a` and `b` of circuits shown in figure is A. `((E_1+E_2)/(C_1+C_2))C_2`B. `((E_2-E_2)/(C_1+C_2))C_2`C. `((E_1+E_2)/(C_1+C_2))C_1`D. `((E_1-E_2)/(C_1+C_2))C_1`

A parallel air capacitance `245muF` has a charge of magnitude `0.48muC` on each plate. The plates are `0.328 mm` apart. (a) What is the potential difference between the plates? (b) What is the area of each plate?(c) What is the surface charge density on each plate?

Find charges on different capacitors.

Three capacitors of capacitances 6μF each are available. The minimum and maximum capacitances, which m&y be obtained are (a) 6μF, 18μF (b) 3μF, 12μF (c) 2μF, 12μF (d) 2μF, 18μF.

Two metal spheres of capacitances C1 and C2, carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy(a) Q1/Q2 < C1/C2 (b) Q1/Q2 = C1/C2(c) Q1/Q2 > C1/c2(d) Q1/Q2 = C2/C1.