A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is
A
4.5 $$\times$$ 10$$-$$6 J
B
3.25 $$\times$$ 10$$-$$6 J
C
2.25 $$\times$$ 10$$-$$6 J
D
1.5 $$\times$$ 10$$-$$6 J
2
NEET 2021
MCQ (Single Correct Answer)
+4
-1
A parallel plate capacitor has a uniform electric field $$\overrightarrow E $$ in the space between the plates. If the distance between the plates is 'd' and the area of each plate is 'A', the energy stored in the capacitor is : ($$\varepsilon $$0 = permittivity of free space)
A
$${{{E^2}Ad} \over {{\varepsilon _0}}}$$
B
$${1 \over 2}{\varepsilon _0}{E^2}$$
C
$${\varepsilon _0}EAd$$
D
$${1 \over 2}{\varepsilon _0}{E^2}Ad$$
3
NEET 2021
MCQ (Single Correct Answer)
+4
-1
The equivalent capacitance of the combination shown in the figure is :
A
3C/2
B
3C
C
2C
D
C/2
4
NEET 2020 Phase 1
MCQ (Single Correct Answer)
+4
-1
The capacitance of a parallel plate capacitor with air as medium is 6$$\mu $$F. With the introduction of a dielectric medium, the capacitance become 30 $$\mu $$F The permittivity of the medium is : $$\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}} \right)$$
