hen a battery is connected to a capacitor both plates acquire equal charges why? What will be if plates are of different size?
Ans.- When a capacitor is charged. Its condensing plates acquires charge due to induction. charge produced by induction is equal in magnitude and opposite in nature and it does not depend upon the size of the conductor. Hence same charge will be developed even on different size of plates.
Q. 3. How the potential changes with distance from a charged particle?
Ans.- Potential due to charge Q at r distance point
V = Q/4πεo r
According to this formula.
(i) When Q = +ve, potential will decrease with increase in distance.
(ii) WhenQ = -ve, potential will increase with increase in distance.
click here TO READ chapter 3 CURRENT ELECTRICITY -short Question & Answer
Q. 4. What are polar and non-polar dielectrics?
Ans.- Polar dielectric
(i)Centers of +ve and –ve charge of each molecule of a di-electric are at different points then, it is called polar dielectric
(ii)Dipole moment of each molecule of a dielectric is not zero , then di-electric is called polar dielectric.
Non-polar dielectric
(i)Centers of +ve and –ve charge of each molecule of a dielectric are at same point then, it is called Non-polar dielectric
(ii)Dipole moment of each molecule of a dielectric is zero , then dielectric is called Non-polar dielectric.
Q. 5. How does the energy lose in redistribution of charge?
Ans.-When two charged conductors are connected then charge flows from the conductor of high potential to the conductor of low potential till then the potentials of both conductors become equal.In this process energy losses in the form of sound energy heat energy and light enter.
Let there are two conductors of capacitances C1& C2 respectively and potentials V1 & V2 respectively.
then their total energy before connecting them
U1 = ½ C1V²1 + ½ C2V²2 ...(i)
a/c to Law of conservation of charge.
C1V1 + C2V2 = C1V + C2V
C1V1 + C2V2 = V (C1 + C2)
V = (C1V1 + C2V2) /(C1 + C2) ..(ii)
This is common potential of both conductor after connecting them.
Total energy after connecting them.
U2 = ½ C1V² + ½ C2V²
= ½ V²(C1 + C2)
=½ [(C1V1 + C2V2) /(C1 + C2)]² × (C1 + C2)
U2 = ½ (C1V1 + C2V2 )²/ (C1 + C2)...(iii)
Loss in energy
= U1–U2
=∆U
= ½ [C1C2(V1–V2)²/(C1+C2)]
This is expression for loss in energy due to redistribution of charge.
Ans.- When a capacitor is charged. Its condensing plates acquires charge due to induction. charge produced by induction is equal in magnitude and opposite in nature and it does not depend upon the size of the conductor. Hence same charge will be developed even on different size of plates.
Q. 3. How the potential changes with distance from a charged particle?
Ans.- Potential due to charge Q at r distance point
V = Q/4π
According to this formula.
(i) When Q = +ve, potential will decrease with increase in distance.
(ii) WhenQ = -ve, potential will increase with increase in distance.
click here TO READ chapter 3 CURRENT ELECTRICITY -short Question & Answer
Q. 4. What are polar and non-polar dielectrics?
Ans.- Polar dielectric
(i)Centers of +ve and –ve charge of each molecule of a di-electric are at different points then, it is called polar dielectric
(ii)Dipole moment of each molecule of a dielectric is not zero , then di-electric is called polar dielectric.
Non-polar dielectric
(i)Centers of +ve and –ve charge of each molecule of a dielectric are at same point then, it is called Non-polar dielectric
(ii)Dipole moment of each molecule of a dielectric is zero , then dielectric is called Non-polar dielectric.
Q. 5. How does the energy lose in redistribution of charge?
Ans.-When two charged conductors are connected then charge flows from the conductor of high potential to the conductor of low potential till then the potentials of both conductors become equal.In this process energy losses in the form of sound energy heat energy and light enter.
Let there are two conductors of capacitances C1& C2 respectively and potentials V1 & V2 respectively.
then their total energy before connecting them
U1 = ½ C1V²1 + ½ C2V²2 ...(i)
a/c to Law of conservation of charge.
C1V1 + C2V2 = C1V + C2V
C1V1 + C2V2 = V (C1 + C2)
V = (C1V1 + C2V2) /(C1 + C2) ..(ii)
This is common potential of both conductor after connecting them.
Total energy after connecting them.
U2 = ½ C1V² + ½ C2V²
= ½ V²(C1 + C2)
=½ [(C1V1 + C2V2) /(C1 + C2)]² × (C1 + C2)
U2 = ½ (C1V1 + C2V2 )²/ (C1 + C2)...(iii)
Loss in energy
= U1–U2
=∆U
= ½ [C1C2(V1–V2)²/(C1+C2)]
This is expression for loss in energy due to redistribution of charge.
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