Parallel Plate Capacitor
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Q. Force of attraction between the plates of a parallel plate capacitor (filled with dielectric of constant k) is
- q22ε0Ak
- q2ε0Ak
- q2ε0Ak
- q22ε0A2k
Q. The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
- Proportional to the square root of the distance between the plates.
- linearly proportional to the distance between the plates
- Independent of the distance between the plates.
- Inversely proportional to the distance between the plates.
Q. A capacitor made of two circular plates each of radius 12 cm and separated by 5 cm. The capacitor is being charged by an external source. The charging current is constant and is equal to 0.15 A. The rate of change of potential difference between the plates is
- 8.39×1010 V/s
- 1.65×1011 V/s
- 2.34×1010 V/s
- 1.87×1010 V/s
Q. Two thin dielectric slabs of dielectric constants K1 and K2(K1<K2) are inserted between the plates of a parallel plate capacitor, as shown in the figure below. The variation of electric field E between the plates with distance d as measured from plate P is correctly shown by -
Q. If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant K placed between the capacitor plates, then bound charge qb can be expressed as
- qb=qf(1−1√K)
- qb=qf(1+1√K)
- qb=qf(1+1K)
- qb=qf(1−1K)
Q. In the following figure an isolated charged conductor is shown. The correct statement will be: -
- EA>EB>EC>ED
- EA<EB<EC<ED
- EA=EB=EC=ED
- EB=EC and EA>ED
Q. For the given arrangement, find the value of electric field at C.
- 2×103 N/C
- 3×103 N/C
- 5×103 N/C
- 9×103 N/C
Q. An uncharged conducting large plate is placed as shown in the figure below. Now an electric field E towards right is applied. Find the induced charge density on the right surface of the plate.
- −ε0E
- ε0E
- −2ε0E
- 2ε0E
Q. Find the net electric potential at the point A for the figure shown below
- KQR
- KQ√R2+d2
- KQd
- zero
Q. Consider a simple RC circuit as shown in Figure 1
Process 1 : In the circuit the switch S is closed at t=0 and the capacitor is fully charged to voltage Ve (i.e. charging continues for time T>>RC ). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is Ec.
Process 2 :
Process 1 : In the circuit the switch S is closed at t=0 and the capacitor is fully charged to voltage Ve (i.e. charging continues for time T>>RC ). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is Ec.
Process 2 :
In a different process the voltage is first set to V03 and maintained for a charging time T>>RC. Then the voltage is raised to 2V03 without discharging the capacitor and again maintained for a time T>>RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1.
These two processes are depicted in Figure 2.
In Process 2, total energy dissipated across the resistance ED is
- ED=13(12CV20)
- ED=3(12cv2o)
- ED=12CV2o
- ED=3CV20
Q. Some equipotential surfaces are shown in figure given below. The magnitude of electric field will be
- 500 V/m
- 300 V/m
- 1000 V/m
- 1200 V/m
Q. Three large plates A, B and C are placed parallel to each other and charges are shown in the figure. The charge that appears on left surface of plate B is
- 6 C
- 3 C
- −6 C
- 8 C
Q. A parallel plate capacitor is charged to a certain potential difference. A dielectric slab of thickness 4.8 mm is inserted between the plates and it becomes necessary to increase the distance beteeen the plates by 4 mm to maintain the same potential difference. The dielectric constant of the slab is :
- 2
- 6
- 4
- 3
Q. The distance between the plates of a charged parallel plate capacitor is 5 cm and the electric field inside the plates is 200 V/cm. An uncharged metal plate of same length and width 2 cm is inserted between the capacitor. The voltage across capacitor after insertion of the plate is
- 0
- 400 V
- 600 V
- 100 V
Q. Two non-conducting infinite plane sheets having charges Q and 2Q are placed parallel to each other as shown in figure. The charge distribution on four faces of two plates are also shown. The electric field intensities at three points 1, 2 and 3 are →E1, →E2 and →E3 repectively then the magnitudes of →E1 , →E2 and →E3 are respectively [Take S is face area of plates]
- Zero; Q2ε0S; zero
- 5Q6ε0S;Qε0S;Q3ε0S
- 5Q6ε0S;Q2ε0S;Zero
- Zero; Qε0S; Zero
Q. A parallel plate capacitor with plates of area 1 m2 each, are at a separation 0.1 m. If the electric field between the plates is 100 NC−1, the magnitude of charge on each plate is:
(Take ε0=8.85×10−12C2Nm2)
(Take ε0=8.85×10−12C2Nm2)
- 8.85×10−10 C
- 7.85×10−10 C
- 6.85×10−10 C
- 9.85×10−10 C
Q. A point charge q1 is placed inside cavity. If q1 is at center of cavity , then →E at a point which is at a distance r from the center of cavity (r>r1) , due to the induced charge on the surface of cavity is
- q14πε0r2 away from the center of cavity.
- q14πε0r21 away from the center of cavity.
- q14πε0r2 towards the center of cavity.
- Zero.
Q. A potential barrier of 0.50 V exists across a P-N junction. If the depletion region is 5.0×10−7 m wide, the intensity of the electric field in this region is
- 1.0×106 V/m
- 1.0×105 V/m
- 2.0×105 V/m
- 2.0×106 V/m
Q. A, B and C are three identical, infinitely long , parallel conducting plates placed as shown in the figure. If q1 and q2 be the charges present on the surfaces left of A and right of B then find the value of q1 and q2
- q1=3Q2, q2=−Q2
- q1=−3Q2, q2=Q2
- q1=3Q2, q2=Q2
- q1=3Q2, q2=3Q2
Q. The electric field between the plates of a parallel plate capacitor of capacitance 2.0 μF drops to one third of its initial value in 4.4 μs when the plates are connected by a thin wire. Find the resistance of the wire. (ln3=1.0985)
- 3.0 Ω
- 2.0 Ω
- 4.0 Ω
- 1.0 Ω
Q.
The plates of a parallel-plate capacitor are made of circular discs of radii 5.0 cm each. If the separation between the plates is 1.0 mm, what is the capacitance ?
Q. In the circuit shown in figure, Find energy stored in 4 μF capacitor at the steady state.
- 0.5 mJ
- 0.8 mJ
- 1.2 mJ
- 1.8 mJ
Q. Figure shows, two equipotential surfaces V1 and V2. The component of electric field (in SI units) along x and y direction will be -
- 100^i and 100^j
- −100^i and 100^j
- 100^i and −100^j
- −100^i and −100^j
Q. The capacitance of a cylindrical capacitor of length 72 mm is 2 pF. If the ratio of radii of its outer cylindrical conductor to that of its inner cylindrical conductor is ex, then x is
- 2
- 4
- 5
- 3
Q. In the circuit shown below, the switch is connected to position 1 for a very long time. What is the maximum current flowing through the circuit, when the switch is taken to position 2?
- 1.5 A
- 2.5 A
- 3.5 A
- 4.5 A
Q. A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Q. An air capacitor of capacitance 10 μF is connected to a battery of 10 V. Now the space between its plates is completely filled with an insulator (K=2). The magnitude of induced charge on the dielectric is
- 100 μC
- 50 μC
- 25 μC
- 200 μC
Q. An air capacitor of capacitance 20 μF is connected to a battery of 20 V. Now the space between its plates is completely filled with an insulator (K=2). The magnitude of induced charge on the dielectric is
- 200 μC
- 100 μC
- 50 μC
- 400 μC
Q. In the circuit shown in the figure, find the maximum energy stored on the capacitor. Initially, the capacitor was uncharged.
- 150 μC
- 100 μC
- 50 μC
- Zero
Q. A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x=0 and positive plate is at x=3d. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to 3d :
- The magnitude of the electric field remains the same.
- The direction of the electric field changes continuously
- The electric potential increases continuously.
- The electric potential increases at first, then decreases and again increases.