Physics

Capacitors and Capacitance

189 Questions

Capacitors and capacitance are crucial topics in physics, covering the storage of electric charge and energy. These concepts explore series and parallel combinations, dielectric materials, and capacitive reactance. Questions frequently appear in various competitive engineering and medical entrance exams.

Series and parallel capacitorsParallel plate capacitorsCapacitive reactance formulasDielectrics and permittivitySpherical capacitors

Capacitors and Capacitance Questions

Multiple choice

A dielectric slab of thickness 4 mm is placed between the plates of a parallel plate condenser. When the distance between the plates is increased by 3.5 mm, then the capacity of the condenser remains same. Find the dielectric of the medium.

  1. 8

  2. 30 x 10-6

  3. 2 x 10-6

  4. 1.87

  5. 12.5 x 10-2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initially the capacity, 

C1 = єoA/d Let the distance is increased by x. When a slab of thickness d is introduced, then capacity C2 =  єoA/ (x+d/єr) since C1 = C2 Therefore, єoA/d = єoA/ (x + d/єr) Or d = (x+d/єr) єr = d/(d – x)     = 4 x10-3/(4 x10-3 – 3.5 x10-3)     = 4/0.5 = 8

Multiple choice

In a network of four capacitors where three are connected in series of 5 μF to a 240 V supply, which of the following charges on the fourth capacitor of 5 μF is connected in parallel with the three capacitors?

  1. 12 x 10-4 C

  2. 400 x 10-6 C

  3. 0.020 x 10-6 C

  4. 48 x 106 C

  5. 0.0069 x 10-6 C
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Three capacitors are connected in series, hence their equivalent capacity will be: Cs = C / 3 = 5 / 3. This equivalent series capacity is then connected in parallel with the fourth capacitor, hence the final equivalent capacity will be: Cs + C 4 = 5 / 3 + 5 = 20 / 3 = 6.66 6 μF. Now, charge on C1, C2 and C3 is same as they are in series. It is given by: Q = Cs V = 5 / 3 x 240 = 5 x 80 x 10 - 6 C = 4 x 10-4 C Charge on C4= C4 x V = 5 x 240 = 1200 x 10-6 = 12 x 10-4C.

Multiple choice

The separation between plates is reduced by half and the space between them is of dielectric constant 5 and capacitance 8 pF. Calculate the value of capacitance of parallel plate capacitor when reduced by half.

  1. 1.25 x 10-12 F

  2. 80 pF

  3. 0.8 pF

  4. 8 pF

  5. 0.8 x 1012 F
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Capacitance of parallel plate capacitor with air between the plates is C0 = Є0A/d.  When the separation between the plates is reduced to half, C1 = Є0A/(d / 2) = 2Є0A/d.  Thus, final capacitance is C2 = 10 x 8 pF = 80 pF.

Multiple choice

What will be the new capacity of two plates if

(a) the distance between the plates is doubled (b) a slab of dielectric constant 8 is introduced between the two plates such that the entire space between the plates is filled by the slab where a parallel plate with air as a dielectric has a capacity of 20 μF.

  1. 10, 2.5 μF

  2. 10, 160 μF

  3. 40, 2.5 μF

  4. 40, 160 μF

  5. 40, 320 μF
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

C = 20 x 10-6 F; Ć = ?  C = A ε0K/d C ∝ 1/d ∴ C = C/2 = 10 μF C ∝ K Ć = K C = 20 x 8 = 160 μF

Multiple choice

A parallel plate air capacitor has the potential difference between the plates as 500 V and rectangular plates each of length 20 cm and breadth 10 cm separated by a distance of 2 mm. Which of the following is the electric field intensity between the two plates?

  1. 0.004 x 10-3 m/V

  2. 250 x 103 V/m

  3. 1000 x 10-3 Vm

  4. 0.885 x 10-10 V/m

  5. 1.12 x 1010 m/V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

L = 0.2 m; b = 0.1 m; d = 2 x 10-3 m; V = 5 x 102V, E = ? C = A ε0/d = (L x b) x ε0/ d = 2 x 10-2x 8.85 x 10-12/2 x 10-3 ∴ C = 8.85 x 10-11F Q = C V = 8.85 x 10-11x 5 x 102 = 4.425 x 10-8C  E = V/d = 500 / 2 x 10-3 = 250 x 103 V/m

Multiple choice

A parallel-plate air capacitor of area 25 cm2 and with plates 1 mm apart is charged to a potential of 100 V. Calculate the new energy by separating the new plates by 0.002 m?

  1. 2.2 x 10-7 J

  2. 1.81 x 107 J

  3. 0.55 x 10-7 J

  4. 0.45 x 107 J

  5. 1.21 x 10-14 J
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Energy = ½ CV2 = ½Є0AV2/d = [8.85 x 10-12 x 25 x 10-4 x 104] /2 x 0.001 = 1.1 x 10-7J New plate separation = 0.002 m; potential across plates is still 100 V. New energy = 0.5 x original energy = 0.55 x 10-7 J The difference in energy is explained by the movement of charge in the wires as the capacitor partly discharges to maintain the potential.

Multiple choice

What is the function of a capacitor?

  1. It produces a magnetic field

  2. It detects changes in temp

  3. It stores electric energy

  4. It detects change in light intensity
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A capacitor is an electronic component that stores electrical energy in an electric field. It does not produce magnetic fields or detect environmental changes.