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Sia ? 6 years, 4 months ago
In metals, n i.e., the number of electrons in unit volume of the conductor is almost independent of temperature. However, t i.e., the relaxation time does vary with temperature. When temperature increases, the thermal speed of the electrons increases as well as, the amplitude of vibration of the positive ions inside the metal conductor also increase, about their mean positions. Thus, the collisions between the electrons and the positive metal ions become more frequent and this decreases the relaxation time, t, leading to an increase in the resistivity of the conductor.
Also, the temperature coefficient of metals is positive.
The number of free electrons in a unit volume of the semi-conductor increases exponentially with an increase in the temperature. This more than compensates the small decrease in t, the relaxation time. Also, the temperature coefficient is negative. Thus the resistivity decreases with a temperature increase in semi-conductors.
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Birmingham is the second-most populous city in the United Kingdom, after London, and the most populous city in the English Midlands.
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Gaurav Seth 6 years, 4 months ago
The net electric charge inside the closed cubic surface = 0, as the dipole consists of equal and opposite charges.
The net electric flux through a closed 3-d surface = charge enclosed/epsilon.
So net electric flux = 0.
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A charged particles produced electric field which in turn produces a magnetic force.If there is a change in electromagnetic field that change will be propagated in waves.So photons don't carry any electric field and thus no charge. They are just a messenger.
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When, the object lies beyond the C a real and inverted image II’ is formed between C and F after reflection from the concave mirror.
From similar triangles BOA and BFP
{tex}\frac{{{\text{AB}}}}{{{\text{PB}}}}{\text{ = }}\frac{{{\text{AO}}}}{{\text{P}}}{/tex}
{tex}\frac{{{\text{AB}}}}{{{\text{AO}}}}{\text{ = }}\frac{{{\text{PB}}}}{{{\text{PF}}}}{/tex}.......(i)
From similar triangles AIB and AFP
{tex}\frac{{{\text{AB}}}}{{{\text{PS}}}}{\text{ = }}\frac{{{\text{BI}}}}{{{\text{PF}}}}{/tex}
(or) {tex}\frac{{{\text{AB}}}}{{{\text{BI}}}}{\text{ = }}\frac{{{\text{PA}}}}{{{\text{PF}}}}{/tex}...........(ii)
Adding (i) and (ii)
{tex}\frac{{{\text{AB}}}}{{{\text{AO}}}}{\text{ + }}\frac{{{\text{AB}}}}{{{\text{BI}}}}{\text{ = }}\frac{{{\text{PB}}}}{{{\text{PF}}}}{\text{ + }}\frac{{{\text{PA}}}}{{{\text{PF}}}}{/tex}
={tex}\frac{{{\text{PB + PA}}}}{{{\text{PF}}}}{/tex}={tex}\frac{{{\text{AB}}}}{{{\text{PF}}}}{/tex}
{tex}\frac{{\text{1}}}{{{\text{AO}}}}{\text{ + }}\frac{{\text{1}}}{{{\text{BI}}}}{\text{ = }}\frac{{\text{1}}}{{{\text{PF}}}}{/tex}
Using cartesian sign convention
AO = Object distance -u
BI= Image distance -v
PF = Focal length = -f
{tex}\frac{1}{{ - u}} + \frac{1}{{ - v}} = - \frac{1}{f}{/tex}
{tex}\frac{1}{u} + \frac{1}{v} = \frac{1}{f}{/tex}
This is the mirror formula.
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Gaurav Seth 6 years, 4 months ago
Figure shows the forces acting on hanging weight and the Balloons.
Fe is the repulsive electrostatic force acting on the charged baloons .
Ft is the upthrust experienced by baloons due to bouyancy. Tension forces acting on the strings is also shown in figure.
It can be seen from figure, net vertical component of Tension is balanced by weight of the hanging mass.
Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1)
at He balloon, horizontal component of Tension is balanced by electrostatic force,
Hence we have, Fe = T sinθ .................(2)
Using eqn.(1), Eqn.(2) is written as, Fe =
.........................(3)
where q is the charge on each balloon, m is mass of hanging object, r is the distance between balloon,
g is acceleration due to gravity, ε0 is the permitivity of free space.
we can see from figure, tanθ =
By substituting all the values in eqn.(3), we get q = 5.5×10-7 C
(distance between balloons is assumed as 0.6 m, not 0.6 cm as shown in figure.
0.6 cm or 6 mm distance between balloons, precisely between centres of balloons, is negligibly small distance , practically not possible )
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Bhavya Tiwari 6 years, 4 months ago
1Thank You