JEE Advanced (Single Correct MCQs): Electrostatics - JEE MCQ

The potential at the surface of a hollow or conducting sphere is same as the potential at the centre of the sphere and any point inside the sphere.

The two charges form an electric dipole. If  we take a point M on the X-axis as shown in the figure, then the net electric field is in –X-direction.

∴ Option (a) is incorrect.
If we take a point N on Yaxis, we find net electric field along  + X direction.
The same will be true for any point on Y-axis. (b) is a correct option.

NOTE : For any point on the equatorial line of a dipole, the electric field is opposite to the direction of dipole moment.

(b) W_{∞ – 0} = q (V_∞–V₀) = q (0 – 0) = 0
∴ (c) is incorrect. The direction of dipole moment is from –ve to + ve. Therefore (d) is incorrect.

C and 2C are in parallel to each other.

∴ Resultant capacity = (2C + C)
C_R = 3C
Net potential = 2V – V
V_R = V

Within the capacitor,

where  A = area of each plate  
d = separation between two plate

With the closing of switch  S₃ and S₁ the negative charge on C₂ will attract the positive charge on C₂ thereby maintaining the negative charge on C₁. The negative charge on C₁ will attract the positive charge on C₁. No transfer of charge will take place. Therefore p.d across C₁ and C₂ will be 30 V and 20 V.

Let C₁ = Capacity of capacitor with K₁
C₂ = Capacity of capacitor with K₂
C₃ = Capacity of capacitor with K₃

C₁ and C₂ are in parallel

C_eq and C₃ are in series

But   for single equivalent capacitor

Electric field lines do not form closed loops. Therefore options (a), (b) and (d) are wrong.
Option (c) is correct. There is repulsion between similar charges.

When S is closed, there will be no shifting of negative charge from plate A to B as the charge – q is held by the charge + q. Neither there will be any shifting of charge from B to A.

NOTE :  As we move along the direction of electric field the potential decreases.

∴ V_A > V_B

Initial energy

when x << a then x² is neglected in denominator..

Initially we know that

Electric field everywhere inside the metallic portion of shell is zero.
Hence options (a) and (d) are incorrect.
Electric field lines are always normal to a surface. Hence option (b) is incorrect. Only option (c) represents the correct answer.

Electric field due to P on O is cancelled by electric field due to S on O.

to Q on O is cancelled by electric field due to T and O.
The electric field due to R on O is in the same direction as that of U on O.
Therefore the net electric field is 

The flux through the Gaussian surface is due to the charges inside the Gaussian surface. But the electric field on the Gaussian surface will be due to the charges present in side the Gaussian surface and outside it. It will be due to all the charges.

Figure shows the electric fields due to the sheets 1, 2 and 3 at point P. The direction of electric fields are according to the charge on the sheets (away from positively charge sheet and towards the negatively charged sheet and perpendicular).

The total electric field

When a charge density is given to the inner cylinder, the potential developed at its surface is different from that on the outer cylinder. This is because the potential decreases with distance for a charged conducting cylinder when the point of consideration is outside the cylinder.
But when a charge density is given to the outer cylinder, it will change its potential by the same amount as that of the inner cylinder. Therefore no potential difference will be produced between the cylinders in this case.

When a positive point charge is placed outside a conducting sphere, a rearrangement of charge takes place on the surface.
But the total charge on the sphere is zero as no charge has left or entered the sphere. 

Let us consider a uniformly charged solid sphere without any cavity. Let the charge per unit volume be σ and O be the centre of the sphere. Let us consider a uniformly charged sphere of negative charged density s having its centre at O'. Also let OO' be equal to a.

Let us consider an arbitrary point P in the small sphere.
The electric field due to charge on big sphere

Also the electric field due to small sphere

 The total electric field

Thus electric field will have a finite value which will be uniform.

The charges make an electric dipole. A and B points lie on the equatorial plane of the dipole.
Therefore, potential at A = potential at B = 0

W = q (V_A – V_B) = q × 0 = 0

The electric field due to A and B at O are equal and opposite producing a resultant which is zero. The electric field at O due to C is

∴ Option [A] is not correct. The electric potential at O is

∴ Option [D] is wrong

Potential energy of the system

∴ Option [B] is wrong Magnitude of force between B and C is

Let the level of liquid at an instant of time 't' be x. Then

Also the capacitance can be considered as an equivalent of two capacitances in series such that

The charges on the surfaces of the metallic spheres are shown in the diagram. It is given that the surface charge densities on the outer surfaces of the shells are equal.
Therefore

Q₁ + Q₂ = 4π(2R)²k = 4[4πR²x]

⇒ Q₂ = 4[4πR² x] - Q₁

= 4[4πR²x] -4πR²x = 3[4πR²x]
Also Q₁ + Q₂+ Q₃ = 4π(3R)² x = 9[4πR²x]
∴ Q₃ = 9[4πR²x] - Q₁ - Q₂ = 9[4πR²x] - [4πR²x] -3[4πR²x] = 5[4πR²x]

⇒ Q₁ : Q₂ :Q₃ = 1 :3 :5

From the figure it is clear that the charge enclosed in the cubical surface is 3C + 2C – 7C = –2C. Therefore the electric flux through the cube is

The electrostatic pressure at a point on the surface of a uniformly charged sphere  = 

∴ The force on a hemispherical shell 

When the electric field is on
Force due to electric field = weight

qE = mg

When the electric field is switched off Weight = viscous drag force

This shows that the electric field acts along + x direction and is a constant. The area vector makes an angle of 45° with the electric field. Therefore the electric flux through the shaded portion whose area is

When S and 1 are connected

The 2μF capacitor gets charged. The potential difference across its plates will be V.
The potential energy stored in 2 μF capacitor

When S and 2 are connected The 8mF capacitor also gets charged. During this charging process current flows in the wire and some amount of energy is dissipated as heat. The energy loss is

The percentage of the energy dissipated  

The pattern of field lines shown in option (c) is correct because
(a) a current carrying toroid produces magnetic field lines of such pattern
(b) a changing magnetic field with respect to time in a region perpendicular to the paper produces induced electric field lines of such pattern.

The frequency of SHM performed by wooden block is
  when electric field is switched on, the value of k and m is not affected and therefore the frequency of SHM remains the same. But as an external force QE starts acting on the block towards right, the mean position of SHM shifts towards right by

correct option is (a).
Note : In SHM if a constant additional force is applied then it only shift the quilibrium position and does not change the frequency fo SHM.