A positive charge Q and a negative charge -2Q are placed 2 cm from each other as
ID: 2257789 • Letter: A
Question
A positive charge Q and a negative charge -2Q are placed 2 cm from each other as
shown in the figure.
a. Find the position of the point A (other than infinity) where the net electric field is equal to zero.
b.Draw the point A on the figure
c. Find the position of the point B (other than infinity) where the net electrostatic potential is zero.
d.Draw on the figure the point B
( +Q ).........................2cm......................( -2Q )
question-2)
water:::
melting point (0C)
heat of fusion 333 kJ/kg
boiling point (100 C)
Heat of vaporization 2255 (kJ/kg)
specific heat (cp)(J/kg)
substance
ice 2086
water 4180
steam 1963
a.How much heat is required to changea 0.1 kg block of ice at temperature -5 oC into steam at a
temperature 110 oC?
b.The heat is provided by a 2 ?resistor connected to 10 V (DC)source. Assuming no heat losses,
how long will it take to change the block of ice from -5 oC into steam at 110 oC?
c.All the steam goes into a container of volume V = 0.01m3. When the temperature is110 oC the pressure inside the container is 1.77x10^6Pa. How many water molecules were there in the ice block?
d.What is the average kinetic energy ofa water molecule in the container?
Explanation / Answer
1) a) As, the two charges are of opposite sign, so zero electric field will occur on the line joining the two charges, but not between the two charges and it will be nearer to the charge of leser magnitude. Let it be at a distance d cm from +Q and 2+d cm from -2Q.Then
Q/d^2 = 2Q/(2+d)^2
=> 4+d^2+4d = 2d^2
=> d^2-4d-4=0
=> d= (4+sqrt(16+16))/2 = 2(1+ 1.414) = 5.636 cm
b) ._____________._____.
A<-----5.636cm------->Q<2cm>-2Q
c) As, the charges are opposite in nature a region of zero electrostatic potential will occur between the two charges.Let it be at a distance k cm from Q. Then it is at a distance (2-k)cm from -2Q.
Q/k = 2Q/(2-k)
=> 2-k = 2k
=> k = 2/3 cm = 0.67 cm
d) .________.____________.
Q<--0.67cm-->B<---1.33cm-------->-2Q
2) a) Total heat required = Heat needed to bring ice at -5oC to 0oC + Heat of fusion+Heat needed to bring water at 0oC to water at 100oC + Heat of vaporization + heat needed to bring steam at 100oC to 110oC = 0.1* (2086*5 + 333000 + 4180*100+ 2255000+ 1963*10) J = 3036.06 KJ
b) Let t be the time.
Heat provided = V^2*t/R = 100*t/2=50*t J= Total heat required
50*t = 3036060 => t= 60721.2 seconds = 16.867 hr
c) Temperature =T= 110oC = 383 K
P= 1.77x10^6Pa
V=0.01 m^3
R= 8.31441 J / K mol
By ideal gas equation, PV= nRT => number of moles = n = 1.77x10^6Pa*0.01 m^3/8.31441 J / K mol* 383 K
=> n = 5.5583 mol
d) Average Kinetic energy of a molecule = 1.5*kb * T = 1.5*1.38*10^(-23)*383 J= 7.9281 * 10^(-21) J
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