Electrostatic Potential and Capacitance
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Is electric potential and capacitance class 12 a hard chapter?
Is electric potential and capacitance class 12 a hard chapter?
This chapter covers the concepts of potential, capacitors, and potential energy. It is considered as one of the easy chapter of the class 12 Physics.
It is the farad (F). The name came from Michael Faraday. SI Unit of Capacitance means the ability of a system to store an electric charge. One farad is the capacitance of a device that needs one coulomb of charge to provide one volt potential difference across it. Mathematically, it is represented b
What is electric potential and capacitance class 12 weightage in JEE Main exam?
What is electric potential and capacitance class 12 weightage in JEE Main exam?
In JEE Main Physics, electric potential and capacitance chapter has a weightage of 3% to 6%. You can expect around 1 or 2 questions from this chapter which carries 4 to 8 marks.
What is the weightage of electric potential class 12 in NEET exam?
What is the weightage of electric potential class 12 in NEET exam?
In NEET Physics exam, the chapter electric potential and capacitance carries a weightage of 2% to 5%, which means you can expect one question out of 45 questions.
The ability to do work on a charge is called the electric potential and the ability to store charge is termed as the capacitance.
Two capacitors having C1 and C2 respectively are connected as shown in figure. Initially, capacitor C1 is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C1 is not connected to uncharged capacitor C2 by closing the switch S. The amount of charge on the capacitor C2, after equilibrium, is:
Two capacitors having C1 and C2 respectively are connected as shown in figure. Initially, capacitor C1 is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C1 is not connected to uncharged capacitor C2 by closing the switch S. The amount of charge on the capacitor C2, after equilibrium, is:
After switch ‘S’ is closed
- (1)
Using KVL
&
A parallel plate capacitor has a uniform electric field ' 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: (ε₀ = permittivity of free space)
A parallel plate capacitor has a uniform electric field ' 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: (ε₀ = permittivity of free space)
E = ½CV²
= ½ (ε? A/d)Ed²
= ½ε? E²Ad
Two hollow conducting spheres of radii and have equal charge. The potential would be:
Two hollow conducting spheres of radii and have equal charge. The potential would be:
The circuit shown in the figure consists of a charged capacitor of capacity 3μF and a charge of 30μC. At time t = 0, when the key is closed, the value of current flowing through the 5MΩ resistor is 'x' μA. The value of 'x' to the nearest integer is---------.
The circuit shown in the figure consists of a charged capacitor of capacity 3μF and a charge of 30μC. At time t = 0, when the key is closed, the value of current flowing through the 5MΩ resistor is 'x' μA. The value of 'x' to the nearest integer is---------.
Potential difference across resistor at time t = V = 30/3 = 10V
Current, I = 10 / (5 * 10? ) = 2? A
A parallel plate capacitor has plate area 100m² and plate separation of 10m. The space between the plates is filled up to a thickness 5m with material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF.
The value of ε? = 8.85 × 10?¹² F.m?¹
The value of 'x' to the nearest integer is ----------.
A parallel plate capacitor has plate area 100m² and plate separation of 10m. The space between the plates is filled up to a thickness 5m with material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF.
The value of ε? = 8.85 × 10?¹² F.m?¹
The value of 'x' to the nearest integer is ----------.
1/C = 5/ (ε? × 100) + 5/ (10ε? × 100)
⇒ C = (ε? × 1000) / 55 = (8.85 × 10? ¹² × 1000) / 55 = 1.61 × 10? ¹? F = 161 pF
