Electric Charges and Fields
- A charge Q µC is placed at the centre of a cube, the flux coming out from any surface will be
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Total flux out of all six faces as per Gauss's theorem should be Q × 10-6 ε0 Therefore, flux coming out of each face = Q × 10-6 C 6 ε0 Correct Option: A
Total flux out of all six faces as per Gauss's theorem should be Q × 10-6 ε0 Therefore, flux coming out of each face = Q × 10-6 C 6 ε0
- A charge q is located at the centre of a cube. The electric flux through any face
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Cube has 6 faces. Flux through any face is given by
φ = q = q4π 6ε0 6(4πε0)
Correct Option: C
Cube has 6 faces. Flux through any face is given by
φ = q = q4π 6ε0 6(4πε0)
- A square surface of side L metres is in the plane of the paper. A uniform electric field E→ (volt/m), also in the plane of the paper, is limited only to the lower half of the square surface (see figure). The electric flux in SI units associated with the surface is
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Flux = E→ . A→
E→ is electric field vector & A→ is area vector. Here, angle between E→ and A→ is 90°.
So, E→ . A→ = 0 ; Flux = 0Correct Option: B
Flux = E→ . A→
E→ is electric field vector & A→ is area vector. Here, angle between E→ and A→ is 90°.
So, E→ . A→ = 0 ; Flux = 0
- A hollow cylinder has a charge q coulomb within it. If φ is the electric flux in units of voltmeter associated with the curved surface B, the flux linked with the plane surface A in units of voltmeter will be
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Since φtotal = φA + φB + φC = q ε0
where q is the total charge.
As shown in the figure, flux associated with the curved surface B is φ = φB
Let us assume flux linked with the plane surfaces A and C be
φA = φC = φ'Therefore, q = 2φ' + φB = 2φ' + φ ε0 ⇒ φ' = 1 q - φ 2 ε0 Correct Option: D
Since φtotal = φA + φB + φC = q ε0
where q is the total charge.
As shown in the figure, flux associated with the curved surface B is φ = φB
Let us assume flux linked with the plane surfaces A and C be
φA = φC = φ'Therefore, q = 2φ' + φB = 2φ' + φ ε0 ⇒ φ' = 1 q - φ 2 ε0
- A square surface of side L meter in the plane of the paper is placed in a uniform electric field E (volt/m) acting along the same plane at an angle θ with the horizontal side of the square as shown in Figure. The electric flux linked to the surface, in units of volt. m, is
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Electric flux , φ = EA cos θ , where θ = angle between E and normal to the surface.
θ = π 2
⇒ φ = 0Correct Option: D
Electric flux , φ = EA cos θ , where θ = angle between E and normal to the surface.
θ = π 2
⇒ φ = 0