Electric Charges and Fields
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The electric field at a distance 3R from the centre of a charged conducting spherical shell of 2 radius R is E. The electric field at a distance R from the centre of the sphere is 2
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Electric field at a point inside a charged conducting spherical shell is zero.
Correct Option: B
Electric field at a point inside a charged conducting spherical shell is zero.
- The electric potential V at any point (x, y, z), all in meters in space is given by V = 4x2 volt. The electric field at the point (1, 0, 2) in volt/meter is
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E→ = - dv î + dv ĵ + dv k̂ dx dy dz
= -8 × î volt/meter
∴ E→ = -8î V / mCorrect Option: D
E→ = - dv î + dv ĵ + dv k̂ dx dy dz
= -8 × î volt/meter
∴ E→ = -8î V / m
- An electric dipole of moment ´p´ is placed in an electric field of intensity ´E´. The dipole acquires a position such that the axis of the dipole makes an angle θ with the direction of the field. Assuming that the potential energy of the dipole to be zero when = 90°, the torque and the potential energy of the dipole will respectively be :
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The torque on the dipole is given as
τ = pE sin θ
The potential energy of the dipole in the electric field is given as
U = – pE cos θCorrect Option: A
The torque on the dipole is given as
τ = pE sin θ
The potential energy of the dipole in the electric field is given as
U = – pE cos θ
- An electric dipole of dipole moment p is aligned parallel to a uniform electric field E. The energy required to rotate the dipole by 90° is
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When electric dipole is aligned parallel θ = 0° and the dipole is rotated by 90° i.e., θ = 90°.
Energy required to rotate the dipole W = Uf – Ui = (–pE cos 90°) – (–pE cos 0°)
= pE.Correct Option: C
When electric dipole is aligned parallel θ = 0° and the dipole is rotated by 90° i.e., θ = 90°.
Energy required to rotate the dipole W = Uf – Ui = (–pE cos 90°) – (–pE cos 0°)
= pE.
- Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as :
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From figure tan θ = Fc ; θ mg kq2 = x x2 mg 2ℓ
or x3 ∝ q2 …(1)
or x3/ 2 ∝ q …(2)
Differentiating eq. (1) w.r.t. time3x2 dx ∝ 2q dq but dq is constant dt dt dt
so x2(v) ∝ q Replace q from eq. (2)
x2(v) ∝ x3/2 or v ∝ x–1/2Correct Option: C
From figure tan θ = Fc ; θ mg kq2 = x x2 mg 2ℓ
or x3 ∝ q2 …(1)
or x3/ 2 ∝ q …(2)
Differentiating eq. (1) w.r.t. time3x2 dx ∝ 2q dq but dq is constant dt dt dt
so x2(v) ∝ q Replace q from eq. (2)
x2(v) ∝ x3/2 or v ∝ x–1/2