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. F_{1} and F_{2} are the forces acting through sides of square.
F_{R} is the resultant of F_{1} , F_{2} and the force of repulsion due to charge Q placed at the corner along diagonal of square.
F_{3} is the force acting through diagonal due to charge q at centre. Hence F_{R} balances F_{3 }to get equilibrium .
If charges Q are positive then charge q has to be -ve to balance F_{R} and F_{3.}
Resultant of F_{1} and F_{2} = √2 × { Q^{2} / ( 4πε_{o} a^{2} ) }
F_{R} = { √2 × [ Q^{2} / ( 4πε_{o} a^{2} ) ] } + { [ Q^{2} / ( 4πε_{o} 2a^{2} ) ] }
F_{3} = { (2Qq) / [ ( 4πε_{o} a^{2} ) ] }
By equating F_{R} and F_{3} and considering q is -ve, we get q = (-Q/4) [ 2√2 + 1 ]
-----------------------------------------------------------
Qn.(2)
Force of attraction between two charges act as centripetal force , so that charge -q_{1} moving with speed v is in circular orbit of radius r
we get,
Period of revolution T = ( 2πr / v ), hence we get

----------------------------------------------------------------------
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 F_{2} on unit positive charge due to charge e is given by, F_{2} = e / [ (4πε_{o} ) s^{2} ]
force F_{1} on unit positive charge due to charge 3e is given by, F_{1} = 3e / [ (4πε_{o} ) (r-s)^{2} ]
By equating F_{1} and F_{2} , we get s= r / [ √3 + 1 ]
----------------------------------------------------------------------------
Qn.(4)
reason :- Electric field will be normal at the surface of conductor

----------------------------------------------------
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
-------------------------------------------------
Qn.(6).
Electric flux φ through hemisphere ( curved surface + base )
By Gauss law , we have, because there are no enclosed charges in hemisphere

where subscript B stands for surface integral over base and subscript CS stands for surface integral over curved surface

------------------------------------------------------------
Qn.(7)
Vertical displacement h of charge is given by, h = (1/2) a t^{2} , 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 t^{2}
Hence vertical deflection is directly proportional to q/m . Hence particle-c has higher charge to mass ratio
------------------------------------------------------------
Qn. (8)

F = q^{2} / (4πε_{o} r^{2} )
if q = 1 C , r = 1000 m
F = 1 × 9 × 10^{9} × 10^{-6} N = 9 × 10^{3} N
-------------------------------------------------
Qn. (9) E_{A} > E_{B}
reason :- at point A, field lines intensity is more ( or lines are closely spaced ) in comparison with point-B.
-----------------------------------
Qn.(10) I am not able to get proper answer

------------------------
Qn.(11) this question is not readable to me, numbers are not visible properly
------------------------
Qn.12
Electric field intensity E is given by
-------------------------------
Q.13
Electric lines are parallel to surface. Hence electric flux through surface is zero
--------------------------------
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 / [ (4πε_{o}) a] ≠ 0.
Electric fields are shown in figure. Net Electric field = E_{R} - E = 2 cos60 E - E = 0
Hence the answer :- E = 0 , V ≠ 0.

_{1}and F

_{2}are the forces acting through sides of square.

_{R}is the resultant of F

_{1}, F

_{2}and the force of repulsion due to charge Q placed at the corner along diagonal of square.

_{3}is the force acting through diagonal due to charge q at centre. Hence F

_{R}balances F

_{3 }to get equilibrium .

_{R}and F

_{3.}

_{1}and F

_{2}= √2 × { Q

^{2}/ ( 4πε

_{o}a

^{2}) }

_{R}= { √2 × [ Q

^{2}/ ( 4πε

_{o}a

^{2}) ] } + { [ Q

^{2}/ ( 4πε

_{o}2a

^{2}) ] }

_{3}= { (2Qq) / [ ( 4πε

_{o}a

^{2}) ] }

_{R}and F

_{3}and considering q is -ve, we get q = (-Q/4) [ 2√2 + 1 ]

_{1}moving with speed v is in circular orbit of radius r

_{2}on unit positive charge due to charge e is given by, F

_{2}= e / [ (4πε

_{o}) s

^{2}]

_{1}on unit positive charge due to charge 3e is given by, F

_{1}= 3e / [ (4πε

_{o}) (r-s)

^{2}]

_{1}and F

_{2}, we get s= r / [ √3 + 1 ]

------------------------------------------------------------

Qn.(7)

Vertical displacement h of charge is given by, h = (1/2) a t

^{2}, 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 t

^{2}Hence vertical deflection is directly proportional to q/m . Hence particle-c has higher charge to mass ratio

------------------------------------------------------------

Qn. (8)

F = q

^{2}/ (4πε_{o}r^{2})if q = 1 C , r = 1000 m

F = 1 × 9 × 10

^{9}× 10^{-6}N = 9 × 10^{3}N-------------------------------------------------

Qn. (9) E

_{A}> E_{B}reason :- at point A, field lines intensity is more ( or lines are closely spaced ) in comparison with point-B.

-----------------------------------

Qn.(10) I am not able to get proper answer

------------------------

Qn.(11) this question is not readable to me, numbers are not visible properly

------------------------

Qn.12

Electric field intensity E is given by

-------------------------------

Q.13

Electric lines are parallel to surface. Hence electric flux through surface is zero

--------------------------------

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 / [ (4πε

_{o}) a] ≠ 0.Electric fields are shown in figure. Net Electric field = E

_{R}- E = 2 cos60 E - E = 0Hence the answer :- E = 0 , V ≠ 0.

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

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