New Capacitance in Dielectric
Trending Questions
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- k=k1+k2+k3+3k4
- K=23(k1+k2+k3)+2k4
- 2k=3k1+k2+k3+1k4
- 1k4=1k1+1k2+1k3+32k4
best given by the expression:
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/635004/original_capacitor.png)
- Aε0kd[1+(αd2)2]
- Akε0d[1+(αd2)]
- Aε0kd[1+(α2d2)]
- Akε0d[1+αd]
- 5
- 1.25
- 4
- 2.5
- 2VK+2
- 2VK+2
- 3VK+3
- 3VK+2
What happens when 2 capacitors are connected with:opposite polarity plates connected together and same polarity plates are connected together
A parallel plate capacitor has a plate area of and plate separation of . The space between the plates is filled up to a thickness of with a material of dielectric constant . The resultant capacitance of the system is . The value of . The value of to the nearest integer is _______.
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in:
None of the above
Reduction of charge on the plates and increase of potential across the plates
Decrease in the potential difference across the plates reduction in stored energy, but no change in the charge on the plates
Increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates
Why does increasing voltage decrease capacitance?
- 25 V, 75 V
- 75 V, 25 V
- 20 V, 80 V
- 50 V, 50 V
- 32
- 2
- 16
- 24
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1723297/original_151.png)
- (K+3)C4
- (K+2)C4
- (K+1)C4
- KC4
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When a dielectric slab is introduced between the plates of an isolated charged capacitor, it
All of the above
Increase the capacitance of the capacitor
Decreases the electric filed between the plates
Decreases the amount of energy stored in the capacitor
0.55
0.45
1.10
2.20
Find the ratio of resistances of two copper rods X and Y of lengths 30 cm and 10 cm respectively and having radii 2 cm and 1 cm respectively.
The dielectric constant of air is
- CV2(K+1K−1)
- CV2(K−1K+1)
- 2CV(K−1K+1)
- 2CV(K+1K−1)
The capacity of a parallel plate condenser is . When a glass plate is
placed between the plates of the conductor, its potential becomes 1/8thof
the original value. The value of dielectric constant will be
1.6
5
8
40
- 30μF
- 120μF
- 40μF
- 160μF
A potential difference is set up between the plates of a parallel plate capacitor by a battery and then the battery is removed. If the distance between the plates is decreased, then how the (a) charge (b) potential difference (c) electric field (d) energy and (e) energy density will change?
A parallel plate capacitor is charged by connecting its plates to the terminals of a battery. The battery is then disconnected and a glass plate is interposed between the plates of the capacitor, then
The capacity increases while the potential difference increases
The capacity increases while the potential difference decreases
capacity decreases while the potential difference decreases
The capacity decreases while the potential difference increases
When a dielectric slab is inserted between the plates of one of the two identical capacitors shown in the figure, then
match the following.
Column−IColumn−II(A) Charge on A(p) increases(B) Potential difference across A(q) decreases(C) Potential difference across B(r) remains constant(D) Charge on B(s) will change
- undefined
- undefined
- undefined
- undefined
If the potential difference between the plates of a charged and isolated parallel-plate condenser of capacitance 1 μF changes from 12 V to 4 V when a dielectric slab of dielectric constant K is introduced to fill the entire space between the plates, then
The change in the charge on the capacitor is 8 µC
The electric field in the capacitor increases by 100 %
The change in the capacity is 3 µF
The value of K is 3
- 12
- 24
- 6
- 3