In a conducting sphere, excess charge distributes itself on the outer surface due to mutual repulsion among like charges. This occurs because charges in a conductor move freely and arrange themselves to minimize repulsive forces, ensuring zero electric field inside the conductor. Answer: (d) On theRead more
In a conducting sphere, excess charge distributes itself on the outer surface due to mutual repulsion among like charges. This occurs because charges in a conductor move freely and arrange themselves to minimize repulsive forces, ensuring zero electric field inside the conductor. Answer: (d) On the outer surface of the sphere.
Φ = Qenc/ε0 where Qenc is the total enclosed charge. Given charges: +5Q,−3Q,+2Q,−4Q Total enclosed charge: Qenc = 5Q−3Q+2Q−4Q = 0 Since the net charge is zero, the outgoing flux is zero. Answer: (b) Zero. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-1/
Φ = Qenc/ε0
where Qenc is the total enclosed charge.
Given charges: +5Q,−3Q,+2Q,−4Q
Total enclosed charge: Qenc = 5Q−3Q+2Q−4Q = 0
Since the net charge is zero, the outgoing flux is zero.
Answer: (b) Zero.
For the system to be in equilibrium, the net force on each charge must be zero. By applying Coulomb’s law and balancing forces symmetrically, the required charge Q at the center must be -q/4 to counteract the repulsion between the two q charges. Answer: (d) -q/4. For more visit here: https://www.tiwRead more
For the system to be in equilibrium, the net force on each charge must be zero. By applying Coulomb’s law and balancing forces symmetrically, the required charge Q at the center must be -q/4 to counteract the repulsion between the two q charges. Answer: (d) -q/4.
The resultant electric field is zero where the fields due to +16q and -4q cancel each other. By using Coulomb’s law and setting the field magnitudes equal, solving for distance gives x = 4L from the charge +16q (toward the right). Answer: (c) 4L. For more visit here: https://www.tiwariacademy.com/ncRead more
The resultant electric field is zero where the fields due to +16q and -4q cancel each other. By using Coulomb’s law and setting the field magnitudes equal, solving for distance gives x = 4L from the charge +16q (toward the right). Answer: (c) 4L.
A vector quantity has both magnitude and direction. Among the given options, electric field (E) is a vector because it has a specific magnitude and direction at every point in space. In contrast, electric charge, electric flux, and electric potential are scalar quantities. Answer: (c) Electric fieldRead more
A vector quantity has both magnitude and direction. Among the given options, electric field (E) is a vector because it has a specific magnitude and direction at every point in space. In contrast, electric charge, electric flux, and electric potential are scalar quantities. Answer: (c) Electric field.
The equation q1 + q2 = 0 signifies an electric dipole because it represents two equal and opposite charges (q1 and 2q) whose sum is zero. An electric dipole consists of such charge pairs separated by a distance, creating an electric field. Answer: (c) Electric dipole. For more visit here: https://wRead more
The equation q1 + q2 = 0 signifies an electric dipole because it represents two equal and opposite charges (q1 and 2q) whose sum is zero. An electric dipole consists of such charge pairs separated by a distance, creating an electric field. Answer: (c) Electric dipole.
The surface integral of the electric field represents electric flux, given by Φ = ∮ E · dA. The SI unit of the electric field (E) is N C⁻¹ and that of area (A) is m², so the unit of electric flux is N·m²·C⁻¹. Answer: (c) Nm² C⁻¹. For more visit here: https://www.tiwariacademy.com/ncert-solutions/claRead more
The surface integral of the electric field represents electric flux, given by Φ = ∮ E · dA. The SI unit of the electric field (E) is N C⁻¹ and that of area (A) is m², so the unit of electric flux is N·m²·C⁻¹. Answer: (c) Nm² C⁻¹.
The charge on a body is always an integral multiple of ±e (electronic charge) due to the quantization of charge. This principle states that charge exists in discrete packets and cannot be divided indefinitely. It is a fundamental property of nature. Answer: (b) Quantisation of charge. For more visitRead more
The charge on a body is always an integral multiple of ±e (electronic charge) due to the quantization of charge. This principle states that charge exists in discrete packets and cannot be divided indefinitely. It is a fundamental property of nature. Answer: (b) Quantisation of charge.
The electric field due to a single point charge is spherically symmetric because the field radiates equally in all directions from the charge. The field strength depends only on the distance from the charge, not the direction. Answer: (c) spherically symmetric. For more visit here: https://www.tiwarRead more
The electric field due to a single point charge is spherically symmetric because the field radiates equally in all directions from the charge. The field strength depends only on the distance from the charge, not the direction.
Answer: (c) spherically symmetric.
For a short electric dipole, the electric field intensity on the axial line is twice that on the equatorial line at the same distance. This follows from the standard dipole field equations: E_axial = (2kP) / r³ and E_equatorial = (kP) / r³, giving E_axial / E_equatorial = 2. Answer: (a) 2. For moreRead more
For a short electric dipole, the electric field intensity on the axial line is twice that on the equatorial line at the same distance. This follows from the standard dipole field equations:
E_axial = (2kP) / r³ and E_equatorial = (kP) / r³, giving E_axial / E_equatorial = 2.
Answer: (a) 2.
Charge on conductiong sphere resides
In a conducting sphere, excess charge distributes itself on the outer surface due to mutual repulsion among like charges. This occurs because charges in a conductor move freely and arrange themselves to minimize repulsive forces, ensuring zero electric field inside the conductor. Answer: (d) On theRead more
In a conducting sphere, excess charge distributes itself on the outer surface due to mutual repulsion among like charges. This occurs because charges in a conductor move freely and arrange themselves to minimize repulsive forces, ensuring zero electric field inside the conductor. Answer: (d) On the outer surface of the sphere.
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Four charges + 5Q,-3Q, + 2Q,-4Q are enclosed inside some surface. What will be the outgoing flux through the surface
Φ = Qenc/ε0 where Qenc is the total enclosed charge. Given charges: +5Q,−3Q,+2Q,−4Q Total enclosed charge: Qenc = 5Q−3Q+2Q−4Q = 0 Since the net charge is zero, the outgoing flux is zero. Answer: (b) Zero. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-1/
Φ = Qenc/ε0
where Qenc is the total enclosed charge.
Given charges: +5Q,−3Q,+2Q,−4Q
Total enclosed charge: Qenc = 5Q−3Q+2Q−4Q = 0
Since the net charge is zero, the outgoing flux is zero.
Answer: (b) Zero.
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A charge Q is placed at the centre of the line joining two charges q and q. The system of three charges will be in equilibrium if Q is
For the system to be in equilibrium, the net force on each charge must be zero. By applying Coulomb’s law and balancing forces symmetrically, the required charge Q at the center must be -q/4 to counteract the repulsion between the two q charges. Answer: (d) -q/4. For more visit here: https://www.tiwRead more
For the system to be in equilibrium, the net force on each charge must be zero. By applying Coulomb’s law and balancing forces symmetrically, the required charge Q at the center must be -q/4 to counteract the repulsion between the two q charges. Answer: (d) -q/4.
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Two point charges + 16 q and – 4 q are located at x = 0 and x = L The location of the point on x-axis at which resultant electric field due to these charges is zero is
The resultant electric field is zero where the fields due to +16q and -4q cancel each other. By using Coulomb’s law and setting the field magnitudes equal, solving for distance gives x = 4L from the charge +16q (toward the right). Answer: (c) 4L. For more visit here: https://www.tiwariacademy.com/ncRead more
The resultant electric field is zero where the fields due to +16q and -4q cancel each other. By using Coulomb’s law and setting the field magnitudes equal, solving for distance gives x = 4L from the charge +16q (toward the right). Answer: (c) 4L.
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Which of the following is a vector quantity?
A vector quantity has both magnitude and direction. Among the given options, electric field (E) is a vector because it has a specific magnitude and direction at every point in space. In contrast, electric charge, electric flux, and electric potential are scalar quantities. Answer: (c) Electric fieldRead more
A vector quantity has both magnitude and direction. Among the given options, electric field (E) is a vector because it has a specific magnitude and direction at every point in space. In contrast, electric charge, electric flux, and electric potential are scalar quantities. Answer: (c) Electric field.
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What does q1 + q2 = 0 signify?
The equation q1 + q2 = 0 signifies an electric dipole because it represents two equal and opposite charges (q1 and 2q) whose sum is zero. An electric dipole consists of such charge pairs separated by a distance, creating an electric field. Answer: (c) Electric dipole. For more visit here: https://wRead more
The equation q1 + q2 = 0 signifies an electric dipole because it represents two equal and opposite charges (q1 and 2q) whose sum is zero. An electric dipole consists of such charge pairs separated by a distance, creating an electric field. Answer: (c) Electric dipole.
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SI unit of surface integral of electric field is
The surface integral of the electric field represents electric flux, given by Φ = ∮ E · dA. The SI unit of the electric field (E) is N C⁻¹ and that of area (A) is m², so the unit of electric flux is N·m²·C⁻¹. Answer: (c) Nm² C⁻¹. For more visit here: https://www.tiwariacademy.com/ncert-solutions/claRead more
The surface integral of the electric field represents electric flux, given by Φ = ∮ E · dA. The SI unit of the electric field (E) is N C⁻¹ and that of area (A) is m², so the unit of electric flux is N·m²·C⁻¹. Answer: (c) Nm² C⁻¹.
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Charge on a body is integral multiple of ± e where e is electronic charge. This is because of
The charge on a body is always an integral multiple of ±e (electronic charge) due to the quantization of charge. This principle states that charge exists in discrete packets and cannot be divided indefinitely. It is a fundamental property of nature. Answer: (b) Quantisation of charge. For more visitRead more
The charge on a body is always an integral multiple of ±e (electronic charge) due to the quantization of charge. This principle states that charge exists in discrete packets and cannot be divided indefinitely. It is a fundamental property of nature. Answer: (b) Quantisation of charge.
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Electric field due to a single charge is
The electric field due to a single point charge is spherically symmetric because the field radiates equally in all directions from the charge. The field strength depends only on the distance from the charge, not the direction. Answer: (c) spherically symmetric. For more visit here: https://www.tiwarRead more
The electric field due to a single point charge is spherically symmetric because the field radiates equally in all directions from the charge. The field strength depends only on the distance from the charge, not the direction.
Answer: (c) spherically symmetric.
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At a given distance from the centre of short electric dipole, field intensity on axial line is k times the field intensity on equatorial line, where k =
For a short electric dipole, the electric field intensity on the axial line is twice that on the equatorial line at the same distance. This follows from the standard dipole field equations: E_axial = (2kP) / r³ and E_equatorial = (kP) / r³, giving E_axial / E_equatorial = 2. Answer: (a) 2. For moreRead more
For a short electric dipole, the electric field intensity on the axial line is twice that on the equatorial line at the same distance. This follows from the standard dipole field equations:
E_axial = (2kP) / r³ and E_equatorial = (kP) / r³, giving E_axial / E_equatorial = 2.
Answer: (a) 2.
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