Q: Four identical thin, square metal sheets, S_1, S_2, S_3 and S_4, each of side a are kept parallel to each other with equal distance d( a) between them, as shown in the figure. Let C_0= _0 a^2 / d, whe

For P All are in series $ gathered C_ e q = C_0 3 \\ P (3) gathered $ For Q $ gathered C_ e q = C_0 2 \\ Q (2) gathered $ For R $ gathered C_ e q = 2 C 3 \\ R (4) gathered $ For S $ aligned & C_ e q =3 C_0 \\ & S (1) aligned $

Q: A container has a base of 50 cm × 5 cm and height 50 cm , as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm × 50

Height of the liquid column = volume base area ⇒ h = 250 cm 3 s - 1 × 10 s 50 cm × 5 cm = 10 cm Now capacitance of the upper part can be written as, C 1 = A 1 ε 0 d = 0 . 50 - 0 . 10 × 0 . 50 × 9 × 10 - 12 5 × 10 - 2 = 0 . 36 × 10 – 10 F Capacitance for the lower part can be written as, C 2 = K A 2 ε 0 d = 3 × 0 . 10 × 0 . 5 × 9 × 10 - 12 5 × 10 - 2 &#8658

Q: In the following circuit C 1 = 12 μF , C 2 = C 3 = 4 μF and C 4 = C 5 = 2 μF . The charge stored in C 3 is μC .

From circuit given, Capacitors C 2 , C 3 , C 4 and C 5 are in parallel connection with constant potential difference as 2 V . Now, charge stored in C 3 is Q 3 = 2 × 4 μC = 8 μC .

Q: A medium having dielectric constant K > 1 fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is d . The capacitor is c

Capacitance of both capacitors are C a = K ε 0 A d and C b = ε 0 A d + d K = ε 0 A K d K + 1 Now, charge stored is q a = K ε 0 A d V and q b = ε 0 A K V d K + 1 . Electric field between each parallel plate capacitor is E a = q a K A ε 0 = K ε 0 A V d K ε 0 A = V d and E b dielectric = E air K = q b K A ε 0 = ε 0 A K V d K + 1 K A ε 0 = V d K + 1 So, from above relations, we can see that capacit

Q: In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q_ 1 C . Then S is switched to position Q . After a long time, the charg

When S is connected to P , In steady state, i = 2 − 1 2 + 1 ⇒ i = 1 3 A Now, Charge on capacitor, q = C V ⇒ q = 1 μF 4 3 V ⇒ q = 4 3 μC ⇒ q = 1.33 μC

Q: In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q_ 1 C . Then S is switched to position Q . After a long time, the charg

When S is connected to Q , i = 2 1 + 2 i = 2 3 A Now, after distribution of potential, q 2 = C V = 1 μF 2 3 V q 2 = 2 3 μC q 2 = 0.67 μC

Question 1

Passage: A container of height 2 m, length 2 m and breadth 1 m is made of insulating vertical walls and two large area horizontal metal plates (M_1 and M_2) which extend far beyond the vertical walls

Accepted Answer

Let A be the area of each chamber. A = 1 1 = 1 m ^2. Distance between plates is d = 2 m . At t = 0, the left chamber is completely filled with liquid ( _r = 15) and right chamber is empty ( _r = 1). Initial capacitance C(0) = C_1 + C_2 = 15 _0 A d + _0 A d = 15 _0 (1) 2 + _0 (1) 2 = 8 _0. Let y_1 and y_2 be the liquid levels in the left and right chambers at time t. By conservation of volume, y_1 + y_2 = 2 m . Let h = y_1 - y_2. The velocity of efflux from the left chamber is v = 2gh . Rate of f

Question 2

Four identical thin, square metal sheets, S_1, S_2, S_3 and S_4, each of side a are kept parallel to each other with equal distance d( a) between them, as shown in the figure. Let C_0= _0 a^2 / d, whe

Accepted Answer

For P All are in series $ gathered C_ e q = C_0 3 \ P (3) gathered $ For Q $ gathered C_ e q = C_0 2 \ Q (2) gathered $ For R $ gathered C_ e q = 2 C 3 \ R (4) gathered $ For S $ aligned & C_ e q =3 C_0 \ & S (1) aligned $

Question 3

A container has a base of 50 &#160; cm &#215; 5 &#160; cm and height 50 &#160; cm , as shown in the figure. It has two parallel electrically conducting walls each of area 50 &#160; cm &#215; 50 &#160;

Accepted Answer

Height of the liquid column  = volume base&#160;area &#8658; h = 250 &#160; cm 3 &#160; s - 1 &#215; 10 &#160; s 50 &#160; cm &#215; 5 &#160; cm = 10 &#160; cm Now capacitance of the upper part can be written as, C 1 = A 1 &#949; 0 d = 0 . 50 - 0 . 10 &#215; 0 . 50 &#215; 9 &#215; 10 - 12 5 &#215; 10 - 2 = 0 . 36 &#215; 10 &#8211; 10 &#160; F Capacitance for the lower part can be written as, C 2 = K A 2 &#949; 0 d = 3 &#215; 0 . 10 &#215; 0 . 5 &#215; 9 &#215; 10 - 12 5 &#215; 10 - 2 &#8658

Question 4

In the following circuit C 1 = 12 &#160; &#956;F , C 2 = C 3 = 4 &#160; &#956;F and C 4 = C 5 = 2 &#160; &#956;F . The charge stored in C 3 is  &#956;C .

Accepted Answer

From circuit given, Capacitors  C 2 , &#160; C 3 , &#160; C 4  and  C 5  are in parallel connection with constant potential difference as  2 &#160; V . Now, charge stored in  C 3  is  Q 3 = 2 &#215; 4 &#160; &#956;C = 8 &#160; &#956;C .

Question 5

A medium having dielectric constant K > 1  fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is d . The capacitor is c

Accepted Answer

Capacitance of both capacitors are  C a = K &#949; 0 A d  and  C b = &#949; 0 A d + d K = &#949; 0 A K d K + 1 Now, charge stored is  q a = K &#949; 0 A d V  and  q b = &#949; 0 A K V d K + 1 . Electric field between each parallel plate capacitor is  E a = q a K A &#949; 0 = K &#949; 0 A V d K &#949; 0 A = V d  and E b dielectric = E air K = q b K A &#949; 0 = &#949; 0 A K V d K + 1 K A &#949; 0 = V d K + 1 So, from above relations, we can see that capacit

Question 6

In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q_ 1 C . Then S is switched to position Q . After a long time, the charg

Accepted Answer

When  S  is connected to P ,  In steady state, i = 2 &#8722; 1 2 + 1 &#8658; i = 1 3 &#160; A Now, Charge on capacitor, q = C V &#8658; q = 1 &#956;F 4 3 &#160; V &#8658; q = 4 3 &#956;C &#8658; q = 1.33 &#160; &#956;C

Question 7

In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q_ 1 C . Then S is switched to position Q . After a long time, the charg

Accepted Answer

When  S  is connected to  Q , i = 2 1 + 2 i = 2 3 &#160; A Now, after distribution of potential, q 2 = C V = 1 &#956;F 2 3 &#160; V q 2 = 2 3 &#160; &#956;C q 2 = 0.67 &#160; &#956;C

Capacitance — Previous Year Questions & Solutions