Test: Electric Field (NCERT) - NEET MCQ

The force per unit charge is known as: E= Fq ​
This formula f is hold for electric field.

In physics, concept of field is a model used to explain the influence that a massive body or charged particle extends into the space around itself, producing a force on another massive body or charged body placed in space. Thus, concept of field is used to explain gravitational and electrostatic phenomena.

Electric field at a point is continuous if there is no charge at that point. And the field is discontinuous if there is charge at that point. So both options (b) and (c) are correct.

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

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
 

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

In figure the field is upward. So the negatively charged electron experiences a downward force. 

∴ The acceleration of electron is
a_c = eE/m_e
The time required by the electron to fall through a distance h is

Here, m = 3.2 x 10^-27 kg, e = 1.6 x 10^-19 C, g = 9.8 m s^-2

Here, n = 10, E = 3.65 x 10⁴ N C^-1
ρ_oil = 1.26gcm^-3 = 1.26 x 10³ kg m^-3
As the droplet is stationary weight of droplet = force due to electric field
or (4/3)πr³ρg = neE ∴ r³ = 3neE/4πρg

or r = (1.13 x 10^-18)^1/3= 1.04 x 10^-6 m

The electric field at the centre of pentagon would be zero.

When a charge q from corner A is removed, electric field at O is E₁ = (q/4πε₀r²) along OA

If the charge q at A is replaced by -q, it is equivalent to adding charge -2q at A. Therefore, electric field at O, E₂ = (2q/4πε₀r²) along OA.

Particles A and B have negative charges because they are being deflected towards the positive plate of the electrostatic field. Particle C has positive charge because it is being deflected towards the negative plate.
∴ Deflection of charged particle in time t in y-direction
h = 0 x t + (1/2)at² = (1/2)(qE/m)t² i.e., h ∝ q/m
As the particle C suffers maximum deflection in y-direction, so it has highest charge to mass q/m ratio.