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
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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).
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
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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).
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
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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.
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
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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.
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
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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).
Assertion (A): The whole charge of a conductor cannot be transferred to another conductor.
Reason (R): The total transfer of charge from one to another is not possible.
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
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Both (A) and (R) are false. Charge can be completely transferred between conductors, for example, by induction or conduction, if placed inside a hollow conductor and connected.
Assertion (A): At a point in space, the electric field points toward east. In the region, surrounding this point the potential will be constant along north and south.
Reason (R): Electric field at a point in space is proportional to rate of change of potential with distance.
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
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Assertion (A) is true because equipotential surfaces are perpendicular to electric field lines. Reason (R) is true as \(E = -\frac{dV}{dr}\). However, (R) describes the relation, but not why potential is constant along north-south specifically for an eastward field.
Assertion (A): A point charge is brought in an electric field. The field at a nearby point will increases, whatever be the nature of charge.
Reason (R): The direction of electric field lines is independent of the nature of charge.
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
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Both (A) and (R) are false. The resultant electric field can increase, decrease or cancel depending on vector sum. Field line direction depends on charge sign.
The electrostatic potential on the surface of a charged solid conducting sphere is 100 volts. Two statements are made in this regard :
Assertion (A): At any point inside the sphere, electrostatic potential is 100 volt.
Reason (R): At any point inside the sphere, electric field is zero.
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
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Concept: Properties of charged conductors.
Principle: Inside a conductor, \( \vec{E} = 0 \) and \( V = \text{constant} \).
Solution: For a solid conducting sphere, potential is uniform throughout its volume and equals surface potential (A is true). This is because the electric field inside a conductor is zero (R is true), implying constant potential. Thus, (R) correctly explains (A).
Assertion (A): If electric field in x-y plane is given by \( \vec{E} = y\hat{i} + x\hat{j} \) then equipotential curve is given by \( xy = \text{ constant} \).
Reason (R): Electric field may not be perpendicular to equipotential surface/curve/line.
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
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Concept: Relation between electric field and equipotential surfaces.
Formula: \( \vec{E} = -\nabla V \).
Solution: From \( \vec{E} = -\nabla V \) for \( \vec{E} = y\hat{i} + x\hat{j} \), potential is \( V = -xy + C \), so equipotential curves are \( xy = \text{ constant} \) (A is true). Electric field lines are always perpendicular to equipotential surfaces (R is false).