MCQ
A parallel plate capacitor is charged to a certain voltage. Now, if the dielectric material (with dielectric constant k) is removed then the
  • A
    Capacitance increases by a factor of k
  • B
    Electric field reduces by a factor k
  • C
    Voltage across the capacitor decreases by a factor k
  • D
    None of these

Answer

  1. None of these

Explanation:

As the capacitor is charged by using cell so potential as well as filed between the plates become constant.

For removing dielectric the capacitance becomes C/ k. Thus capacitance decreases by a factor of k.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The material suitable for making electromagnets should have:
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
Assertion (A): The kinetic energy of a charged particle moving in a uniform magnetic field does not change. Reason (R): In a uniform magnetic field no force acts on the charge particle.

The diffraction effect can be observed in

(a) Only sound waves

(b) Only light waves

(c) Only ultrasonic waves 

(d) Sound as well as light waves

A small cylindrical soft iron piece is kept in a galvanometer so that

(a) A radial uniform magnetic field is produced

(b) A uniform magnetic field is produced

(c) There is a steady deflection of the coil

(d) All of these

The distance between the two charges +q  and -q  of a dipole is  r. On the axial line at a distance d from the centre of dipole, the intensity is proportional to

(a)  

(b)

(c)

(d)

A steady electric current is flowing through a cylindrical conductor.

A radio transmitter operates at a frequency of 880 kHz and a power of 10 kW. The number of photons emitted per second are

(a) 1.72  

(b) 1327   

(c) 13.27   

(d) 0.075   

The resistance of a galvanometer is 50 ohms and the current required to give full scale deflection is 100μA. In order to convert it into an ammeter, reading upto 10A, it is necessary to put a resistance of

(a) 5  in parallel

(b) 5  in parallel

(c)  in series

(d) 99,950  in series

A coil of resistance 10 W and an inductance 5H is connected to a 100 volt battery. Then energy stored in the coil is

(a) 125 erg

(b) 125 J

(c) 250 erg

(d) 250 J

For a cell of e.m.f. 2V, a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a  2Ω resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is

(a) 0.25 Ω

(b) 0.50 Ω

(c) 0.80 Ω

(d) 1.00 Ω