- ADecreases
- BRemains unchanged
- CBecomes zero
- DIncreases
Explanation:
In general capacitance of parallel plate capacitor is given by:
$\text{C}=\frac{\text{K}\in_0\text{A}}{\text{d}}$
Where C is capacitance, k is relative permittivity of dielectric material, ϵ0 is permittivity of free space constant, A is area of plates and d is distance between them. Therefore, the capacitance of parallel plates is increased by the insertion of a dielectric material. Further, the capacitance is inversely proportional to the electric field between the plates, and hence the presence of the dielectric decreases the effective electric field.
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A silver and a copper voltmeters are connected in parallel across a 6 volt battery of negligible resistance. In half an hour, 1 gm of copper and 2 gm of silver are deposited. The rate at which energy is supplied by the battery will approximately be (Given E.C.E. of copper = 3.294 and E.C.E. of silver = 1.118
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(a) 64 W |
(b) 32 W |
(c) 96 W |
(d) 16 W |
In a working nuclear reactor, Cadmium rods, (control rods) are used to
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(a) Speed up neutrons |
(b) Slow down neutrons |
|
(c) Absorb some neutrons |
(d) Absorb all neutrons |
Two thin lenses, one of focal length + 60 cm and the other of focal length – 20 cm are put in contact. The combined focal length is
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(a) + 15 cm |
(b) – 15 cm |
(c) + 30 cm |
(d) –30 cm |
The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is
|
(a) |
(b) |
(c) |
(d) |
$\text{v}=\frac{\text{v}_1+\text{v}_2}{2}$
$\text{v}=\sqrt{\text{v}_1\text{v}_2}$
$\frac{2}{\text{v}}=\frac{1}{\text{v}_1}+\frac{1}{\text{v}_2}$
$\frac{1}{\text{v}}=\frac{1}{\text{v}_1}+\frac{1}{\text{v}_2}$
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|
(a) |
(b) 4C |
(c) |
(d) 2C |
Activity of radioactive element decreased to one third of original activity
in 9 years. After further 9 years, its activity will be
|
(a) |
(b) |
(c) |
(d) |