Test: Electric Charges and Fields - 2 - CUET Humanities MCQ

Let us consider the forces on a ball, at point Q.
Three forces act on it:
(i) Tension T in the thread
(ii) Force mg due to gravity
(iii) Force F due to coulomb repulsion along the +ve x-direction
For equilibrium, the sum of the x and y components of these forces must be zero.
i.e. T cos 60° - F = 0
And T sin 60° - mg = 0
So, we have
F = mg cot 60° = (√3/10) x 10^-3 10 x (1/√3) = 10^-3 N
Now,

Putting F = 10^-3 N, r = 0.3 m and  we get q = 10^-7 C.

Hence the correct choice is 3. 

Let the electric field be zero at a distance x from charge 4q. Then

Or 2(r – x) = x
Or x = 2r/3, which is option (4).

The charge inside the closed surface is given by
q = net electric flux pass through the surface × ε₀
= (4 × 10³ - 8 × 10³) ε₀
Therefore, q = -4 × 10³ ε₀ coulomb.

Given: v_x = 10 ms^–1
Since the electric field is directed along the y–axis, the acceleration of the body along the y–direction is

Therefore, the velocity of the body along the y-axis at time t = 10 s is
V_y = at = 1 x 10 = 10 ms^-1
∴ Magnitude of the resultant velocity (or we can say speed),
v = 
= 
Hence, the correct option is (3).

The rod attracts positive charge to the top electrode, creating an excess of negative charge at the leaves. The leaves having the same charge, are repelled and spread out.

The electric field due to an extremely short dipole at a distance r can be written as

Thus, .

The charge at Q exerts an attractive force F on charge at P along PQ. The charge at R exerts a repulsive force on charge at P along PS of magnitude F. The angle between these two forces is 120°. From the parallelogram law, the magnitude of the resultant force is


Or F_r = F
As shown in the figure, the direction of the resultant force is along the negative x-direction.
Hence, the correct option is (2).

If all the charges are in equilibrium, then the system is also in equilibrium.
Consider all the forces acting at the point B:
F_A = k towards AB
F_C = k towards CB
F_D =  towards DB
F_O =  towards BO

Force at B away from the centre = F_AC + F_D
= 
Force at B towards the centre = F_O = 
For equilibrium of charge at B, F_AC + F_D = F_O

Electric field due to charge –Q on the shell at a distance r from its centre is (for r > R):

Electric field due to charge +Q at the centre at a distance r is(for r<R)

∴ Net electric field E (for r > R) = E₁ – E₂ = 0 (because distance R of any point from both the charges will become same and charges defeats the effect of each other)
Only option 2 and 3 satisfied this condition.
Now,For r < R, the electric field due to the shell charge (-Q) in the cell is zero and the electric field due to charge +Q at the centre decreases as 1/r².
Hence, the correct graph is (3).

Let Q be the displaced position of the pith ball.
At this position, the following forces act on the ball:
(i) Weight (W = mg) of the ball in the vertical downward

(ii) Force (F = qE) due to the applied electric field along PQ
(iii) Tension (T) in the string along QO
From ΔOPQ,
PQ/OP = F/W = qE/W
Or 
(Here, q = 1 nC = 10^–9 C; m = 0.5 g = 0.5  10^–3 kg)
 E = 
= 1.96 x 10⁵ N C^–1
tan θ = 2/50 = 0.04
Or θ = 2°18'

Opposite charges attract each other; thus, the body can be either positively or negatively charged.

For the positive charge, electric field lines start from the charge and ends at infinity, so the given charge is positive.

When two plastic straws are rubbed with a silk cloth, they will acquire similar charge and therefore, repel each other. Now, as the silk cloth will have opposite charge as compared to the plastic straws, the plastic straws will attract the silk cloth.

Electroscope is based on the principle that like charges repel each other. If a charged rod is brought near electroscope, the leaves will move apart on account that like charges repel each other whether the charges are positive or negative.