Sir/Maa'm,

Its my request to solve this tommorow itself its very urgent

as my test approaches so early

i need answer in picture form so i could take printout

question attached below

regards,

Asked by rahmanafeef | 28th Mar, 2020, 10:50: PM

Expert Answer:

Qn.(1)
Forces on central charge are getting balanced as shown in fig.2 , because equal forces acting on opposite direction.
 
Forces acting on the charges placed at corner are shown in fig.3. F1 and F2 are the forces acting through sides of square.
 
FR is the resultant of F1 ,  F2 and the force of repulsion due to charge Q placed at the corner along diagonal of square.
 
F3 is the force acting through diagonal due to charge q at centre. Hence FR balances F3 to get equilibrium .
 
If charges Q are positive then charge q has to be -ve to balance FR and F3.
 
Resultant of F1 and F2 = √2 × { Q2 / ( 4πεo a2 ) }
 
FR = { √2 × [ Q2 / ( 4πεo a2 ) ] } + {  [ Q2 / ( 4πεo 2a2 ) ] }
 
F3 = { (2Qq) / [ ( 4πεo a2 ) ]  }
 
By equating FR and F3 and considering q is -ve, we get q = (-Q/4) [ 2√2 + 1 ]
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Qn.(2)
Force of attraction between two charges act as centripetal force , so that charge -q1 moving with speed v is in circular orbit of radius r
 
begin mathsize 14px style fraction numerator q subscript 1 space q subscript 2 over denominator 4 pi epsilon subscript o r squared end fraction space equals space fraction numerator m space v squared over denominator r end fraction end style
we get, begin mathsize 14px style v space equals space square root of fraction numerator q subscript 1 space q subscript 2 over denominator m space left parenthesis 4 pi epsilon subscript o right parenthesis space r end fraction end root end style
Period of revolution T = ( 2πr / v ), hence we get

begin mathsize 14px style T space equals space square root of fraction numerator 16 pi squared epsilon subscript o space m space r cubed over denominator q subscript 1 space q subscript 2 end fraction end root end style
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Qn.(3)
Let the point charge be placed at a distance s from charge e. Hence its distance from charge 3e is (r-s)
 
force F2 on unit positive charge due to charge e  is given by,  F2 = e / [ (4πεo ) s2 ]
 
force F1 on unit positive charge due to charge 3e is given by,  F1 = 3e / [ (4πεo ) (r-s)2 ]
 
By equating F1 and F2 , we get  s= r / [ √3 + 1 ]
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Qn.(4)
reason :-  Electric field will be normal at the surface of conductor
 
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Qn.(5)
An electric dipole in a uniform electric field experiences maximum moment of couple
when the dipole is placed perpendicular to the direction of the field 
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Qn.(6).
Electric flux φ through hemisphere ( curved surface + base ) begin mathsize 14px style surface integral E. d a end style
By Gauss law , we have, begin mathsize 14px style surface integral E. d a space equals space 0 end style because there are no enclosed charges in hemisphere
 
begin mathsize 14px style surface integral E. d a space equals space integral integral subscript B space E. d a space plus space integral integral subscript C S end subscript E. d a space equals space 0 end style
where subscript B stands for surface integral over base and subscript CS stands for surface integral over curved surface

begin mathsize 14px style integral integral subscript C S end subscript E. d a space equals space minus integral integral subscript B space E. d a space space equals space pi R squared E end style
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Qn.(7)
Vertical displacement h of charge is given by,  h = (1/2) a t2 , where a = qE/m , is acceleration experienced by charge q which is having mass m.
E is electric field intensity
 
hence vertical deflection, h = (1/2) (q/m) E t2
 
Hence vertical deflection is directly proportional to q/m .  Hence particle-c has higher charge to mass ratio
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Qn. (8)

F = q2 /  (4πεo r2 )
 
if q = 1 C , r = 1000 m
 
F = 1 × 9 × 109 × 10-6 N = 9 × 103 N
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Qn. (9)  EA > EB
reason :- at point A, field lines intensity is more ( or lines are closely spaced ) in comparison with point-B.
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Qn.(10) I am not able to get proper answer
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Qn.(11) this question is not readable to me, numbers are not visible properly
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Qn.12
Electric field intensity E is given by
begin mathsize 14px style E space equals space integral fraction numerator lambda space d l over denominator 4 pi epsilon subscript o R squared end fraction space equals space fraction numerator lambda R over denominator 4 pi epsilon subscript o R squared end fraction integral subscript negative pi divided by 2 end subscript superscript pi divided by 2 end superscript d theta space equals space fraction numerator lambda over denominator 4 epsilon subscript o R end fraction end style
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Q.13
Electric lines are parallel to surface. Hence electric flux through surface is zero
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Q.14
Point O is at equal distance a from all the charges.
 
Potential is scalar function of distance a . Hence potential due to each charge is getting added together and
 
total potential V = 3Q / [ (4πεo) a]  ≠ 0.
 
Electric fields are shown in figure. Net Electric field = ER - E = 2 cos60 E - E = 0
 
Hence the answer :- E = 0 , V ≠ 0.
 

Answered by Thiyagarajan K | 29th Mar, 2020, 12:36: PM

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