Electrostatics - NEET Physics Questions
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Electrostatics

Question 181: easy

Assertion (A): We cannot produce electric field in a neutral conductor.


Reason (R): Neutral conductor cannot produce electric field.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

In electrostatic equilibrium, the electric field inside a conductor is zero due to charge redistribution. A neutral conductor has no net charge, so it cannot be a source of electric field. Both assertion (A) and reason (R) are true, but (R) does not correctly explain (A); the zero field inside is due to charge mobility and redistribution, not simply its neutrality.

Question 182: easy

Assertion (A): In a given situation of arrangement of charges, an additional charge is placed outside the Gaussian surface. In this situation, in the Gauss theorem \(\oint \vec{E}.d\vec{s} = \frac{q_{in}}{\epsilon_0}\) remains unchanged whereas electric field \(vec{E}\) is changed.


Reason (R): Electric field \(\vec{E}\) at any point on the Gaussian surface is due to inside charge only.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true. An external charge does not change the net charge enclosed by the Gaussian surface \(q_{in}\), so the total electric flux \(\oint \vec{E}.d\vec{s}\) remains unchanged as per Gauss's Law. However, the electric field \(\vec{E}\) at any point on the surface is the vector sum of fields from all charges, both inside and outside, so it will change. Reason (R) is false because the electric field at any point is due to both internal and external charges.

Question 183: easy

Assertion (A): Angular momentum of the two dipole system is not conserved.


Reason (R): There is a net torque on the system.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

For an isolated system of two dipoles, the torques they exert on each other are internal forces. Internal torques within a system cancel each other out, leading to zero net torque on the system. Therefore, according to the conservation of angular momentum                            \(\frac{d\vec{L}}{dt} = \vec{\tau}_{ext}\)), if the net external torque is zero, the angular momentum of the system is conserved. Hence, both assertion (A) and reason (R) are false.

Question 184: easy

Assertion (A): Continuity equation explains conservation of electric charge.


Reason (R): Gauss law shows diversion when inverse square law is not obeyed.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Continuity equation describes conservation of charge. Gauss's law is a fundamental law valid irrespective of the inverse square law and does not show 'diversion' based on its obedience. Thus, (A) is true, (R) is false.

Question 185: easy

Assertion (A): A moving charge particle may gets energy from electric field.


Reason (R): Electric field works on moving charge.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

An electric field exerts a force \( \vec{F} = q\vec{E} \) on a charge `\( q \)`. If the charge moves, work \( W = \int \vec{F} \cdot d\vec{l} \) can be done, changing its energy. Hence, both are true and (R) explains (A).

Question 186: easy

Assertion (A): Electric field intensity at surface of a uniformly charged spherical shell is `\( E \)`. If shell is punctured at a point then intensity at punctured point becomes `\( E/2 \)`.


Reason (R): Electric field intensity due to a spherical charge distribution can be found out by using Gauss law.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

The field at a puncture is `\( E/2 \)` due to superposition. Gauss's law helps find the field for symmetric distributions, but it doesn't explain the `\( E/2 \)` effect at the puncture directly. Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Question 187: easy

Assertion (A): If X-ray is allowed to fall on uncharged gold leaf in evacuated glass chamber of electroscope, leaves will diverge.


Reason (R): Uncharged gold leaves will get charged positively when x-ray falls on it.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

X-rays have sufficient energy to eject electrons from the gold leaves (photoelectric effect), leaving them positively charged. The like positive charges on the leaves cause them to repel and diverge. Thus, (R) correctly explains (A).

Question 188: easy

Assertion (A): When a dipole is placed in a non- uniform electric field dipole must experience non zero force and torque.


Reason (R): Electric dipole is in stable equilibrium in non uniform electric field.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

In a non-uniform field, `\( +q \)` and `\( -q \)` experience different forces, leading to a net force and torque. Thus (A) is true. However, a dipole is generally not in stable equilibrium in a non-uniform field, so (R) is false.

Question 189: easy

Assertion (A): When charges are shared between two bodies, there occurs no loss of charge, but there does occur a loss of energy.


Reason (R): In case of sharing of charges energy of conservation fails.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Total charge is always conserved during sharing. Electrical potential energy may decrease and convert to heat, implying 'loss' of electrical energy (A is true). However, total energy is always conserved; (R) is false because energy conservation never fails.

Question 190: easy

Assertion (A): Excess charge on a conductor resides entirely on the outer surface.


Reason (R): Like charges repel each other.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true. Excess charges on a conductor spread out on the outer surface due to mutual repulsion. Reason (R) is also true, as like charges repel. (R) correctly explains (A).