Electric field lines always start from a positive charge and end at a negative charge. In case of a single charge, electric lines of force start from positive charge and end at infinity.
The magnitude of a charge depends on the number of lines of force emanating from a charge, i.e. higher the number of lines of force, higher the magnitude of charge and vice-versa.
(i) In the given figure, the electric lines of force emanate from A and C. Therefore, charges A and C must be positive.
(ii) The number of electric lines of force emanating is maximum from charge C here, so C must have the largest magnitude.
(iii) Point between two like charges, where electrostatic force is zero, is called neutral point. So, the neutral point lies between A and C only.
Now, the position of neutral point depends on the strength of the forces of charges. Here, more number of electric lines of force show higher strength of charge C than A. So, neutral point lies near A.

(i) Here, the direction of magnetic field is perpendicularly inwards to the plane of paper. If a wire of irregular shape turns into a circular shape, then its area increases (\(\because \) the circular loop has greater area than the loop of irregular shape) so that, the magnetic flux linked also increases. Now, the induced current is produced in a direction such that it decreases the magnetic field [i.e. the current will flow in such a direction, so that the wire forming the loop is pulled inward in all directions (to decrease the area)], i.e current is in anti-clockwise direction, ie. along adcba.  
(ii) When a circular loop deforms into a narrow straight wire, the magnetic flux linked with it also decreases. The current induced due to change in flux will flow in such a direction that it will oppose the decrease in magnetic flux, so it will flow anti-clockwise, i.e along adcba due to which the magnetic field produced will be out of the plane of paper. 

Let the distance between the plates be increased by a very small distance\(\triangle x\) = Force x Increased distance
= F.\(\Delta x\)     ...(i)  

Increase in volume of capacitor
= Area of plates x Increased distance
= A.\(\Delta x\)
u = Energy density = \(\frac { Energy }{ Volume } \)
Increase in energy = u x voulme = u.A.\(\Delta x\)   .....(ii)
As, Energy = Work done
F.\(\Delta x\) = u.A.\(\Delta x\) \([\because From \ Eqs.(i) \ and \ (ii)]\) 
 \(\Rightarrow\) F = u.A
\(=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 }.A\quad [\because u=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 } \ and \ E=\frac { V }{ d } ]\)
\(=\frac { 1 }{ 2 } \varepsilon _{ 0 }.\frac { { V }^{ 2 } }{ { d }^{ 2 } } .A=\left( \frac { { \varepsilon }_{ 0 }A }{ d } .A \right) \frac { V }{ d } \times \frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 2 } .E.C.V=\frac { 1 }{ 2 } QE \ [\because C=\frac { { \varepsilon }_{ 0 }A }{ d } ,CV=Q]\)

(a) The point O is equidistant from all the charges at the end points of pentagon.
(b) When charge q is removed from A electric field at O
  E = \(\frac { q\times 1 }{ 4\pi \varepsilon _{ \circ }r^{ 2 } } \)
(c) If charge q at A is replaced by -q then it add chage -2q so electric field become E = \(\frac { 2q }{ 4\pi \varepsilon _{ \circ }r^{ 2 } } \)
(ii) when pentagon is replaced by n-side regular polygon with charge q at each of its corners, the electric field at O would continue to be zero as symmetricity of the charge. 

(From the given figure we can analyse that the chrolorine atom is at the centre of the cube. Thus due to symmetry the force due to all Cs atoms on CI atom will cancel out.)
E = F/q'
F = 0
E = 0
(ii) net force on CI atom at A would be given by
F = \(\frac { e^{ 2 } }{ 4\pi \varepsilon _{ \circ }r^{ 2 } } \)
\(\\ r=\sqrt { (0.20)^{ 2 }+(0.20)^{ 2 }+(0.20)^{ 2 }\times 10^{ -9 } } m\)
\( =0.346\times 10^{ -9 }\ m\)
\(\\ F=\frac { e^{ 2 } }{ 4\pi \varepsilon _{ \circ }r^{ 2 } } \)
\( =1.92\times 10^{ -9 }\ N\) 

According to Ampere's circuital law, the net field is given by \(B={ \mu }_{ 0 }nI\)
(a) The net magnetic field is given by
\({ B }_{ net }={ B }_{ 2 }-{ B }_{ 1 }-{ \mu }_{ 0{ n }_{ 2 } }I-{ \mu }_{ 0 }{ n }_{ 1 }\quad [\because { I }_{ 2 }={ I }_{ 1 }=I]\)
 \(={ \mu }_{ 0 }I({ n }_{ 2 }-{ n }_{ 1 })\)
The direction is from B to A.
(b) As the magnetic fields due to S₁ is confined solely inside S₁ as the solenoids are assumed to be very long. So, there is no magnetic field outside S₁  due to current in S_1, similarly, there is no field outside S₂ .
B_net = 0

(i) Here, in the figure, the electric lines of force emanate from A and C. Therefore, charges A and C must be positive.
(ii) The number of electric lines of force emanating is maximum from charge C here, so C must have the largest magnitude.
(iii) Point between two like charges where electrostatic force is called neutral point. So, the neutral point lies between A and C only.
Now, the position of neutral point depends on the strength of the forces of charges. Here more number of electric lines of force show higher strength of charge C than A. So, neutral point lies near A.

(a)Electrostatic field lines cannot start from a negative charge.
(b)Electrostatic field lines cannot end at positive charge
(c)Electrostatic field lines cannot form closed loops.

Given, m = 9.1 x 10^-31 kg, q = 1.6 x 10^-19 C
\(E={ 10 }^{ 4 }{ V\diagup m,v={ 10 }^{ 6 } }m\diagup s,t=3ms=0.003s\)
As, \(y=\frac { qE }{ 2m{ v }^{ 2 } } { x }^{ 2 }\)
 \(y=\frac { qE }{ 2m } \left( \frac { x }{ v } \right) ^{ 2 }=\frac { qE }{ 2m } { t }^{ 2 }\left[ \therefore time(t)=\frac { distance(x) }{ speed(v) } \right] \)
 \(y=\frac { \left( 1.6\times { 10 }^{ -19 } \right) \times \left( { 10 }^{ 4 } \right) \times \left( 0.003 \right) ^{ 2 } }{ 2\times \left( 9.1\times { 10 }^{ -31 } \right) } =7.9\times { 10 }^{ 9 }m\)

\(Here,length,2a=2cm=2\times { 10 }^{ -2 }m.\)
  \(\theta ={ 60 }^{ 0 },\tau =8\sqrt { 3 } N-m\)
Charges, Q = 4\(\times\) 10^-9 C, U = ?
As we know that, \(\tau =Q(2a) E\sin { \theta } \)
\(\Longrightarrow \)Electric field,
\(e=\frac { \tau }{ Q(2a)\sin { \theta } } =\frac { 8\sqrt { 3 } }{ 4\times { 10 }^{ -9 }\times 2\times { 10 }^{ -2 }\times \sin { { 60 }^{ 0 } } } N/C\)
\(\\ \therefore Potential\quad energy, U=-\rho E\cos { \theta } =-Q(2a)E\cos { \theta }\)
\( \\ =-4\times { 10 }^{ -9 }+2\times { 10 }^{ -2 }\times \frac { 8\sqrt { 3 } \times \cos { { 60 }^{ 0 } } }{ 4\times { 10 }^{ -9 }\times 2\times { 10 }^{ -2 }\times \sin { { 60 }^{ 0 } } }\)
\( \\ \Longrightarrow U=-\frac { 8\sqrt { 3 } }{ \sqrt { 3 } } =-8J\)​​​​​​