A parallel plate capacitor with air between the plates, has a capacitance of 2 pF. The plate separation is doubled and a dielectric slab is inserted so that it completely fills the space between the plates. The capacitance of the capacitor becomes 4 pF. The dielectric constant of the slab is:

A parallel plate capacitor with air between the plates, has a capacita

1.	A parallel plate capacitor with air between the plates, has a capacitance of 2 pF. The plate separation is doubled and a dielectric slab is inserted so that it completely fills the space between the plates. The capacitance of the capacitor becomes 4 pF. The dielectric constant of the slab is:
A.	0.5
B.	2
C.	4
D.	8
Answer» D. 8

Discussion

No Comment Found

Related MCQs

Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ(r) = kr, where r is the distance from the centre. Two charges A and B, of -Q each, are placed on diametrically opposite points, at equal distance, ‘a’ from the centre. If A and B do not experience any force, then:
In the above circuit, \(C = \sqrt {\frac{3}{2}} \;\mu F,{R_2} = 20\Omega ,\;L = \frac{{\sqrt 3 }}{{10}}H\) and R1 = 10 Ω Current in L – R1 path is I1 and in C - R2 path is I2. The voltage of AC source is given by \(V = 200\sqrt 2 \;sin\left( {100t} \right)\) difference between I1 and I2 is
A positive charge +q is placed at the centre of a hollow metallic sphere of inner radius a and outer radius b. the electric field at a distance r from the centre is denoted by E. In this regards, which one of the following statement is correct?
If a free electron moves through a potential difference of 1 kV, then the energy gained by the electron is given by
A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is \(\left( {{\rm{Take,\;}}{{\rm{\varepsilon }}_0} = 8.85 \times {{10}^{ - 12}}\frac{{{{\rm{C}}^2}}}{{{\rm{N}} - {{\rm{m}}^2}}}} \right)\)
A parallel plate capacitor with air between the plates, has a capacitance of 2 pF. The plate separation is doubled and a dielectric slab is inserted so that it completely fills the space between the plates. The capacitance of the capacitor becomes 4 pF. The dielectric constant of the slab is:
Four point charges -q, +q, +q and -q are placed on y-axis at y = -2d, y = -d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as:
Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle as shown in the figure.
A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d

A parallel plate capacitor with air between the plates, has a capacitance of 2 pF. The plate separation is doubled and a dielectric slab is inserted so that it completely fills the space between the plates. The capacitance of the capacitor becomes 4 pF. The dielectric constant of the slab is:

Discussion

No Comment Found

Related MCQs

Reply to Comment