Electrostatic Potential and Capacitance

More Topic from Physics

Question 1

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be

  • 1

  • 2

  • 1/4

  • 1/2

Solution

D.

1/2

Question 2

A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :

  • 24

  • 32

  • 2

  • 16

Solution

B.

32

To hold 1 KV potential difference minimum four capacitors are required in series
⇒ C1 = 1/4
for one series.
So for Ceq to be 2μF, 8 parallel combinations are required.

⇒ Minimum no. of capacitors = 8 × 4 = 32

Question 3

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other tow corners. If the net electrical force on Q is zero, then Q/q equals

  • negative 2 square root of 2
  • 1

  • -1

  • negative fraction numerator 1 over denominator square root of 2 end fraction

Solution

A.

negative 2 square root of 2

Three forces F41, F42 and f43 acting on Q are shown Resultant of F41 + F43
space equals space square root of 2 space straight F subscript each
space equals space square root of 2 space fraction numerator 1 over denominator 4 πε subscript straight o end fraction Qq over straight d squared
Resultant on Q becomes zero only when ‘q’ charges are of negative nature.

straight F subscript 4 comma 2 end subscript space equals space fraction numerator 1 over denominator 4 space πε subscript straight o end fraction fraction numerator straight Q space straight x space straight Q over denominator left parenthesis square root of 2 straight d right parenthesis squared end fraction
rightwards double arrow space square root of 2 dQ over straight d squared space equals space fraction numerator straight Q space xQ over denominator 2 straight d squared end fraction
rightwards double arrow space square root of 2 space straight x space straight q space equals space fraction numerator straight Q space straight x space straight Q over denominator 2 straight d squared end fraction
therefore space straight q space equals space minus fraction numerator straight Q over denominator 2 square root of 2 end fraction space space or space straight Q over straight q space equals space minus 2 square root of 2
Question 4

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 µF and 9 µF capacitors), at a point distant 30 m from it, would equal:


  • 240N/C

  • 360N/C

  • 420N/C

  • 480N/C

Solution

C.

420N/C

Resultant circuit,

As, charge on 3μF = 3μF x 8V = 24μC
Charge on 3μF = 3μF x 2V = 18 μC
charge on 4μF +Charge on 9μF
= (24 + 18)μC = 42μC
therefore,
Electric field at a point distant 30 m
fraction numerator 9 space straight x space 10 cubed straight x space 42 space straight x space 10 to the power of negative 6 end exponent over denominator 30 space straight x space 30 end fraction space equals space 420 space straight N divided by straight C

Question 5

A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is

  • straight pi over 4 square root of LC
  • 2 straight pi space square root of LC
  • square root of LC
  • straight pi square root of LC

Solution

A.

straight pi over 4 square root of LC

As initially, charge is maximum
∴ q = qocos ωt
Current, 

2