Test: Forces between multiple charges & Electric Field (27 Oct) - JEE MCQ

 As shown in figured raw AD ⊥ BC.



Distance AO of the centroid O from A

∴ Force on Q at O due to charge q₁ = q at A,

Similarly, force on O due to charge q₂ = q at B
 along BO and force on Q due to charge q₃ = q at c

Angle between forces F₂ and F₃ = 120°
By parallelogram law, resultant of  along OA
∴ Total force on Q = 

Forces of repulsion on 1μC charge at O due to 3 μC charge, at A and C are equal and opposite. So they cancel each other.
Similarly, forces of attraction of 1μC charge at O due to -4μC charges at B and D are also equal and opposite. So they also cancel each other.
Hence the net force on the charge of 1μC at O is zero. 

Here, distance of point from the centre of the sphere, r = 20 cm = 0.2m
Electric field, E =  -1.2 x 10³ N C^-1
 

For the system to be in equilibrium, net force on q = 0 or 
or Q = (-q)/4 or Q/q = -(1/4)

Let a charge 2q be placed at P, at a distance l from A where charge q is placed, as shown in figure.
The charge 2q will not experience any force, when  force of repulsion on it due to q is balanced by force of attraction on it due to -3q at B where AB = d
or 
(l + d)² = 3l² or 2l² - 2ld - d² = 0

If charge 2q is placed between A and B then 

F = qE = 5 x 10^-6 x 2 x 10⁵ = 1 N 
Since, the particle is thrown against the field
∴ a = -F/m = -(1/10^-3) = 10³ms^-2
As v² - u² = 2as
∴ 0² - (20)² = 2 x (-10³) x s or s = 0.2 m

As 
If q ' = 2q, then E' =  E' = 2E
So electric field is doubled.

From diagram, force on q₁(= q) at A,


where  is the unit vector along BC 
Force on 
(where   is the unit vector along AC)
Force on q₃(= -q) at C,

where  unit vector along the direction bisecting ∠BCA