According to the superposition principle, the total electric potential at a point is equal to the sum of the potentials due to each charge at that point.
Consider the point at which the total potential zero is located at a distance x from the charge +q as shown in the figure.

The total electric potential at P is zero.
V_tot = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ x } -\frac { 2q }{ (9-x) } \right) \)=0
Which gives \(\frac { q }{ x } -\frac { 2q }{ (9-x) } \)
or \(\frac { 1 }{ x } =\frac { 2 }{ (9-x) } \)
Hence, x = 3m

(a) The work done to arrange the charges in the corners of the square is independent of the way they are arranged. We can follow any order.
(i) First, the charge +q is brought to the corner P. This requires no work since no charge is already present, W_P = 0
(ii) Work required to bring the charge –q to the corner Q = (-q) x potential at a point Q due to +q located at a point P
W_Q = -q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ a } =-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^{ 2 } }{ a } \)
(iii) Work required to bring the charge +q to the corner R = q x potential at the point R due to charges at the point P and Q.
W_R = q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( -\frac { q }{ a } +\frac { q }{ \sqrt { 2 } a } \right) \)
= \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }^{ 2 } }{ a } \left( -1+\frac { 1 }{ \sqrt { 2 } } \right) \)
(iv) Work required to bring the fourth charge –q at the position S = q × potential at the point S due the all the three charges at the point P, Q and R.
W_s = - q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ a } +\frac { q }{ a } -\frac { q }{ \sqrt { 2 } a } \right) \)
W_s = \(-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^2 }{ a } \left( 2-\frac { 1 }{ \sqrt { 2 } } \right) \)
(b) Work required to bring the charge q′ to the center of the square = q′ x potential at the center point O due to all the four charges in the four corners.
The potential created by the two +q charges are canceled by the potential created by the -q charges which are located in the opposite corners. Therefore the net electric potential at the center O due to all the charges in the corners is zero.
Hence no work is required to bring any charge to the point O. Physically this implies that if any charge q′ when brought close to O, then it moves to the point O without any external force.

The electric flux through the rectangular area
\({ \Phi }_{ E }=\vec { E } .\vec { A } \) = EA cosθ = 100 x 5 x 10 x 10^-4 x cos60°
⇒ \({ \Phi }_{ E }\) = 0.25 Nm²C^-1
For θ = 0°
\({ \Phi }_{ E }=\vec { E } .\vec { A } \)= EA
= 100 x 5 x 10 x 10^-4 = 0.5 Nm²C^-1

If the conductor has no net charge, then its motion is the same as usual projectile motion of a mass m which we studied in Kinematics (unit 2, vol-1 XI physics). Here, in this problem, in addition to downward gravitational force, the charge also will experience a downward uniform electrostatic force.
The acceleration of the charged ball due to gravity = -g\(\hat { j } \)
The acceleration of the charged ball due to uniform electric field =\(-\frac { qE }{ m } \hat { j } \)
The total acceleration of charged ball in downward direction \(\vec { a } =-\left( g+\frac { qE }{ m } \right) \vec { j } \)
It is important here to note that the acceleration depends on the mass of the object. Galileo’s conclusion that all objects fall at the same rate towards the Earth is true only in a uniform gravitational field. When a uniform electric field is included, the acceleration of a charged object depends on both mass and charge.
But still the acceleration a = \(\left( g+\frac { qE }{ m } \right) \) is constant throughout the motion. Hence we use kinematic equations to calculate the range, maximum height and time of flight. In fact we can simply replace g by \(g+\frac { qE }{ m } \) in the usual expressions of range, maximum height and time of flight of a projectile.

Note that the time of flight, maximum height, range are all inversely proportional to the acceleration of the object. Since \(\left( g+\frac { qE }{ m } \right) \) >g for charge +q, the quantities T, h_max, and R will decrease when compared to the motion of an object of mass m and zero net charge. Suppose the charge is –q, then \(\left( g-\frac { qE }{ m } \right) \)< g, and the quantities T, h_max and  R will increase. Interestingly the trajectory is still parabolic as shown in the figure.

(a) The capacitance of the capacitor in the presence of dielectric is
C = \(\frac { { \varepsilon }_{ r }{ \varepsilon }_{ 0 }A }{ d } =\frac { 5\times 8.85\times 10^{ -12 }\times 6 \times 10^{-4}}{ 6\times 10^{ -3 } } \)
= 44.25 x 10^-13F = 4.425 pF
The stored charge is
Q = CV = 44.25 x 10^-13 x 10
= 442.5 x 10^-13C = 44.25pC
The stored energy is
\(U=\frac { 1 }{ 2 } \) CV² = \(\frac { 1 }{ 2 } \) x 44.25 x 10^-13 x 100
= 2.21 x 10^-10 J
(b) After the removal of the dielectric, since the battery is already disconnected the total charge will not change. But the potential difference between the plates increases. As a result, the capacitance is decreased.
New capacitance is
C₀=\(\frac { C }{ { \varepsilon }_{ r } } =\frac { 44.25\times 10^{ -12 } }{ 5 } \)
= 0.885 x 10^-12 F = 0.885 pF
The stored charge remains same and 44.25 pC. Hence newly stored energy is
U₀ =\(\frac { { Q }^{ 2 } }{ 2{ C }_{ 0 } } =\frac { { Q }^{ 2 }{ \varepsilon }_{ r } }{ 2C } =\varepsilon _{ r }U\)
= 5 x 2.21 x 10^-10J = 11.05 x 10^-10J
The increased energy is ΔU = (11.05 - 2.21) x 10^-10 J  = 8.84 x 10^-10 J
When the dielectric is removed, it experiences an inward pulling force due to the plates. To remove the dielectric, an external agency has to do work on the dielectric which is stored as additional energy. This is the source for the extra energy 8.84 x 10^–10 J.