Electrostatics MSQ - IIT JAM MCQ

i.e. Net field at A is towards a.

The correct answers are: The ratio of charge of elements a₁ and a₂ is  r₁/r₂, The element a₁ produced greater magnitude of electric field at A than element a₂, The elements a₁ and a₂ produce same potential at A, The direction of net electric field at A is towards element  a₂

Charge is distributed over the surface of conductor in such a way that net field due to this charge and outside charge q is zero inside. Field due to only q is non-zero.

The correct answers are: On the surface of conductor the net charge is negative., On the surface of conductor at some points charges are negative and at some points charges may be positive distributed non uniformly, Inside the conductor electric field due to point charge is non zero

V at origin ≠ 0

 0
 

From origin to r = 2,  field is towards origin.

The correct answers are: The electric field intensity is zero at r = 2m, A positive charged particle disturbed from origin along the x-axis will restore back to origin, Where potential is maximum on the x-axis, field intensity is zero

Increasing the accelerating voltage means increasing speed of the electron, thereby decreasing time spent between the plates. It will reduce x.
Increasing deflecting voltage means increasing electric field between the plates, making acceleration of electron greater. Increasing distance once again will change electric field between the plates.

The correct answers are: increasing the accelerating voltage, increasing the deflecting voltage, increasing the distance between the two deflecting plates

Speed will be maximum when acceleration becomes zero. i.e. when Kx = EQ

By work-energy theorem : 

The correct answers are: The maximum speed of the block will be  The maximum compression of the spring will be 

The correct answers are: the horizontal range of the particle is 10m, the time of flight of the particle is 2sec, the maximum height reached is 5m

The correct answers are: The flux of the electric field through the sphere is zero, The electric field is not zero at any where on the sphere

The distribution of charge on the outer surface, depends only on the charges outside, and it distributes itself such that the net, electric field inside the outer surface due to the charge on outer surface and all the outer charges is zero. Similarly the distribution of charge on the inner surface, depends only on the charges inside the inner surface, and it distributes itself such that the net, electric field outside the inner surface due to the charge on inner surface and all the inner charges is zero.
Also the force on charge inside the cavity is due to the charge on the inner surface.

The correct answers are: only induced charge on inner surface will redistribute., both induced charge on outer and inner surface will redistribute., force on charge q placed inside the cavity will change.

The correct answers are: The electric field calculated by Gauss’s law is the field due to all the charges, The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface

; (As V_a = 0)


The correct answers are: For  potential V_x = 0, For x > a  potential