PHYSICS DOUBTS (Electrostatics)
Q 1: The electric strength of air is 2*10^7 N/C. The maximum charge that a metallic sphere of diameter 6mm can hold is?
a. 3nC b. 20nC c. 1.5nC d. 2nC
PRINTED ANSWER: d. 2nC
SOURCE: Arihant Physics for JEE Main
MY DOUBT: The obvious formula here would be Q=4 pi e0 r^2 E . However, instead of using E = 2*10^7 N/C an additional step is printed: "As maximum value of E = 10% dielectric strength, hence E = 10%(2*10^7 ) = 2*10^6 N/C"
Where did that step come from? Is there any such law? If yes, does it always hold true?
Q 2: A 4microF capacitor is charged to 400V. If its plates are joined through a resistance of 4kOhm, then heat produced in the resistance will be?
a. 0.16J b. 0.32J c.0.64J d. 1.28J
PRINTED ANSWER: (none)
SOURCE: MH-CET 2000
MY DOUBT: How do you solve this? Since V=IR does not hold true, I guess it would be right to use the formula for discharging of a capacitor (?)
Q 3: A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of -3Q, the new potential difference between the same two surfaces is?
a. V b. 2V c. 4V d. -2V
PRINTED ANSWER: (none)
SOURCE: IIT JEE - 1989
MY DOUBT: Not a concrete idea as to how to solve this. The situation is similar to that of a capacitor, right? Induced charges?
(That s it, thanks :) )
kvnmurty:
please write one part in each question . it is difficult to answer for long and multi-part questions. Further , it is difficult to moderate questions and answers.
Answers
Answered by
9
1.
Regarding the dielectric strength, for air it seems to be 3 * 10^6 V/m or N/Coulomb. This is listed in many references. Regarding using a fraction like 1/10, actually, a corona discharge starts near sharp points when electric voltage/field nears the breakdown situation. Further impurities in air and easily ionizable gases in air can decrease the effective value of dielectric strength by a factor. This is the real life situation. As the given exercise does not specify these factors, we don't consider them.
Q = 4 π ∈ * r² * E
I guess the value given for E is mistyped, it may be 2 * 10^6 or 3 * 10^6.
2.
Energy required to charge a capacitor of capacitance C to a voltage V and charge Q is given by :
Energy = 1/2 Q V = 1/2 Q²/C = 1/2 C V² , as Q = C V
= 1/2 * 4 * 10^-6 * 400 * 400 = 0.32 Joules
Whatever resistance you use, this energy will be dissipated through the resistance slower or faster. The current initially will be high and it will decrease exponentially and will be zero after a long time.
3. There is some thing not right with the question. Instead of potential difference, it should be ratio of potentials at the outer surface in the two cases.
Let the radius of the inner solid sphere be R1. Let the outer radius of the outer shell be R2. It does not matter what the thickness of the shell is and if there is any gap between the inner sphere and the shell.
The outer surface of the inner sphere has all the charge +Q distributed on its surface. The inner surface of the shell has induced charge of -Q. The outer surface of the shell has the charge +Q distributed equally on the total surface.
Then the potentials V1 and V2 at points P1 and P2 distances R1 and R2 is:
V1 = 1/4πε * Q/R1 and V2 = 1/4πε * (Q-Q+Q)/R2
V2 - V1 = Q/4πε * [1/R2 - 1/R1 ] it is actually negative.
V = - Q/4πε * [R2 - R1] / R1 R2
Now the outer shell is given a charge of -3Q.
So there will be a charge -Q on the inner surface of shell and -2Q will be at the outer surface. Hence, the potentials V1' and V2' at points P1 and P2 will be :
V1' = V1, V2' = 1/4πε * [ Q /R2 -Q /R2 - 2 Q/R2] = -2Q / 4πε R2
V2' - V1' = Q/4πε * [ -2 /R2 - 1/R1 ]
V' = - Q/4πε * [2R1 + R2] / R1 R2
So V'/V = ( 2 R1 + R2 ) / (R2 - R1)
I think the question is wrongly framed. Instead of potential difference, it should be ratios of electric potential at the outer surface of the conductor shell, in two cases. So that is,
V2 ' / V2 = - 2
If that is the case then option d will be right.
Regarding the dielectric strength, for air it seems to be 3 * 10^6 V/m or N/Coulomb. This is listed in many references. Regarding using a fraction like 1/10, actually, a corona discharge starts near sharp points when electric voltage/field nears the breakdown situation. Further impurities in air and easily ionizable gases in air can decrease the effective value of dielectric strength by a factor. This is the real life situation. As the given exercise does not specify these factors, we don't consider them.
Q = 4 π ∈ * r² * E
I guess the value given for E is mistyped, it may be 2 * 10^6 or 3 * 10^6.
2.
Energy required to charge a capacitor of capacitance C to a voltage V and charge Q is given by :
Energy = 1/2 Q V = 1/2 Q²/C = 1/2 C V² , as Q = C V
= 1/2 * 4 * 10^-6 * 400 * 400 = 0.32 Joules
Whatever resistance you use, this energy will be dissipated through the resistance slower or faster. The current initially will be high and it will decrease exponentially and will be zero after a long time.
3. There is some thing not right with the question. Instead of potential difference, it should be ratio of potentials at the outer surface in the two cases.
Let the radius of the inner solid sphere be R1. Let the outer radius of the outer shell be R2. It does not matter what the thickness of the shell is and if there is any gap between the inner sphere and the shell.
The outer surface of the inner sphere has all the charge +Q distributed on its surface. The inner surface of the shell has induced charge of -Q. The outer surface of the shell has the charge +Q distributed equally on the total surface.
Then the potentials V1 and V2 at points P1 and P2 distances R1 and R2 is:
V1 = 1/4πε * Q/R1 and V2 = 1/4πε * (Q-Q+Q)/R2
V2 - V1 = Q/4πε * [1/R2 - 1/R1 ] it is actually negative.
V = - Q/4πε * [R2 - R1] / R1 R2
Now the outer shell is given a charge of -3Q.
So there will be a charge -Q on the inner surface of shell and -2Q will be at the outer surface. Hence, the potentials V1' and V2' at points P1 and P2 will be :
V1' = V1, V2' = 1/4πε * [ Q /R2 -Q /R2 - 2 Q/R2] = -2Q / 4πε R2
V2' - V1' = Q/4πε * [ -2 /R2 - 1/R1 ]
V' = - Q/4πε * [2R1 + R2] / R1 R2
So V'/V = ( 2 R1 + R2 ) / (R2 - R1)
I think the question is wrongly framed. Instead of potential difference, it should be ratios of electric potential at the outer surface of the conductor shell, in two cases. So that is,
V2 ' / V2 = - 2
If that is the case then option d will be right.
click on thank you link and select best answer
please write one part in each question . it is difficult to answer for long and multi-part questions. Further , it is difficult to moderate questions and answers.
Similar questions