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Electrostatics

NEET Physics · 98 questions · Page 6 of 10 · Click an option or "Show Solution" to reveal answer

Q51
Three concentric spherical shells have radii a, b and c (a < b < c) anf have surface charge densities σ\sigma , −-σ\sigma and σ\sigma respectively. If V A , V B and V C denote the potentials of the three shells, then, for c = a + b, we have
A V C = V B ≠ \ne V A
B V C ≠ \ne V B ≠ \ne V A
C V C = V B = V A
D V C = V A ≠ \ne V B
Correct Answer
Option D
Solution
VA=14πε0{qAa+qBb+qCc}{V_A} = {1 \over {4\pi {\varepsilon _0}}}\left\{ {{{qA} \over a} + {{qB} \over b} + {{qC} \over c}} \right\}
=4π4πε0{a2σa−b2σb+c2σc}= {{4\pi } \over {4\pi {\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}
VA=1ε0{a2σa−b2σb+c2σc}{V_A} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}
VB=1ε0{a2σa−b2σb+c2σc}{V_B} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}
VC=1ε0{a2σa−b2σb+c2σc}{V_C} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}

Given c = a + b. If a = a, b = 2a and c = 3a for example, as c > b > a,

VA=1ε0{a2σa−4b2σ2a+c2σc}{V_A} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{4{b^2}\sigma } \over {2a}} + {{{c^2}\sigma } \over c}} \right\}
VB=1ε0{a2σ2a−4a2σ2a+c2σc}{V_B} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over {2a}} - {{4{a^2}\sigma } \over {2a}} + {{{c^2}\sigma } \over c}} \right\}
VC=1ε0{a2σ3a−4a2σ3a+c2σc}{V_C} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over {3a}} - {{4{a^2}\sigma } \over {3a}} + {{{c^2}\sigma } \over c}} \right\}

It can seen by taking out common factors that V A = V C > V B i.e., V A = V C ≠\ne V B

Q52
The electric potential at a point (x, y, z) is given by V = −-x 2 y −- xz 3 + 4 The electric field at that point is
A E→=i^2xy+j^(x2+y2)+k^(3xz−y2)\overrightarrow E = \widehat i2xy + \widehat j\left( {{x^2} + {y^2}} \right) + \widehat k\left( {3xz - {y^2}} \right)
B E→=i^z3+j^xyz+k^z2\overrightarrow E = \widehat i{z^3} + \widehat jxyz + \widehat k{z^2}
C E→=i^(2xy−z3)+j^xy2+k^3z2x\overrightarrow E = \widehat i\left( {2xy - {z^3}} \right) + \widehat jx{y^2} + \widehat k3{z^2}x
D E→=i^(2xy+z3)+j^x2+k^3xz2\overrightarrow E = \widehat i\left( {2xy + {z^3}} \right) + \widehat j{x^2} + \widehat k3x{z^2}
Correct Answer
Option D
Solution

The electric potential at a point, V = –x 2 y – xz 3 + 4. The field

E→=−∇→V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^)\overrightarrow E = - \overrightarrow \nabla V = - \left( {{{\partial V} \over {\partial x}}\widehat i + {{\partial V} \over {\partial y}}\widehat j + {{\partial V} \over {\partial z}}\widehat k} \right)
∴E→=i^(2xy+z3)+j^x2+k^(3xz2)\therefore \overrightarrow E = \widehat i\left( {2xy + {z^3}} \right) + \widehat j{x^2} + \widehat k\left( {3x{z^2}} \right)
Q53
The electric potential at a point in free space due to charge Q coulomb is Q × \times 10 11 volts. The electric field at that point is
A 4πε0Q×10204\pi {\varepsilon _0}Q \times {10^{20}} volt/m
B 12π\pi 0 Q × \times 10 22 volt/m
C 4πε0Q×10224\pi {\varepsilon _0}Q \times {10^{22}} volt/m
D 12πε0Q×102012\pi {\varepsilon _0}Q \times {10^{20}} volt/m
Correct Answer
Option C
Solution
V=14πε0×QRV = {1 \over {4\pi {\varepsilon _0}}} \times {Q \over R}

= Q × 10 11 volt …(i)

E=14πε0×QR2E = {1 \over {4\pi {\varepsilon _0}}} \times {Q \over {{R^2}}}

= V/R = Q × 10 11 ×

4πε0{4\pi {\varepsilon _0}}

× 10 11 from ...(i) =

4πε0{4\pi {\varepsilon _0}}

× Q × 10 22 volt/m

Q54
A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is
A E along KO
B 3E along OK
C 3E along KO
D E along OK
Correct Answer
Option D
Solution

By the symmetry of the figure, the electric fields at O due to the portions AC and BD are equal in magnitude and opposite in direction.

So, they cancel each other.

Similarly, the field at O due to CD and AKB are equal in magnitude but opposite in direction.

Therefore, the electric field at the centre due to the charge on the part ACDB is E along OK.

Q55
A hollow cylinder has a charge q coulomb within it. If ff 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 V-m will be
A q2ε0{q \over {2{\varepsilon _0}}}
B ϕ3{\phi \over 3}
C qε0−ϕ{q \over {{\varepsilon _0}}} - \phi
D 12(qε0−ϕ){1 \over 2}\left( {{q \over {{\varepsilon _0}}} - \phi } \right)
Correct Answer
Option D
Solution

Let

ϕA\phi _A

,

ϕB\phi _B

and

ϕC\phi _C

are the electric flux linked with A, B and C. According to Gauss theorem,

ϕA+ϕB+ϕC\phi _A + \phi _B + \phi _C

=

qε0{q \over {{\varepsilon _0}}}

Since

ϕA\phi _A

=

ϕC\phi _C
⇒2ϕA=qε0−ϕB\Rightarrow 2{\phi _A} = {q \over {{\varepsilon _0}}} - {\phi _B}
⇒2ϕA=qε0−ϕ\Rightarrow 2{\phi _A} = {q \over {{\varepsilon _0}}} - \phi
(Given ϕB=ϕ)\left( {Given\,{\phi _B} = \phi } \right)

∴\therefore

ϕA=12(qε0−ϕ){\phi _A} = {1 \over 2}\left( {{q \over {{\varepsilon _0}}} - \phi } \right)
Q56
Charges +q and −-q are placed at points A and B respectively which are a distance 2L apart, C is the midnight between A and B. The work done in moving a charge + Q along the semicircle CRD is
A qQ2πε0L{{qQ} \over {2\pi {\varepsilon _0}L}}
B qQ6πε0L{{qQ} \over {6\pi {\varepsilon _0}L}}
C −- qQ6πε0L{{qQ} \over {6\pi {\varepsilon _0}L}}
D qQ4πε0L{{qQ} \over {4\pi {\varepsilon _0}L}}
Correct Answer
Option C
Solution

Potential at C = V C = 0 Potential at D = V D

=k(−qL)+kq3L=−23kqL= k\left( {{{ - q} \over L}} \right) + {{kq} \over {3L}} = - {2 \over 3}{{kq} \over L}

Potential difference V D – V C

=−23kqL=14πε0(−23.qL)= - {2 \over 3}{{kq} \over L} = {1 \over {4\pi {\varepsilon _0}}}\left( { - {2 \over 3}.{q \over L}} \right)

⇒\Rightarrow Work done = Q (V D – V C )

=−23×14πε0qQL=−qQ6πε0L= - {2 \over 3} \times {1 \over {4\pi {\varepsilon _0}}}{{qQ} \over L} = {{ - qQ} \over {6\pi {\varepsilon _0}L}}
Q57
Three point charges +q, −- 2q and + q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = aa, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are
A 2qa\sqrt 2 qa along the line joining points (x = 0, y = 0, z = 0) and (x = aa, y = a, z = 0)
B qaa along the line joining points (x = 0, y = 0, z = 0) and (x = aa, y = a, z = 0)
C 2qa\sqrt 2 qa along +x direction
D 2qa\sqrt 2 qa along +y direction.
Correct Answer
Option A
Solution

This consists of two dipoles, –q and +q with dipole moment along with the +y-direction and –q and +q along the x-direction.

∴\therefore The resultant moment =

q2a2+q2a2=2qa\sqrt {{q^2}{a^2} + {q^2}{a^2}} = \sqrt 2 qa

Along the direction 45° that is along OP where P is (+a, +a, 0).

Q58
A square surface of side L metres is in the plane of the paper. A uniform electric field E→\overrightarrow 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 inits associated with the surface is
A EL 2
B EL 2 /2ε\varepsilon 0
C EL 2 /2
D zero
Correct Answer
Option D
Solution

Electric flux,

ϕE=∫E→.dS→{\phi _E} = \int {\overrightarrow E .d\overrightarrow S }
=∫EdScos⁡θ=∫EdScos⁡90o=0= \int {EdS\cos \theta } = \int {EdS\cos {{90}^o} = 0}

The lines are parallel to the surface.

Q59
An electric dipole of moment p→\overrightarrow p is lying along a uniform electric field E→\overrightarrow E . The work done in rotating the dipole by 90 o is
A pE
B 2pE\sqrt 2 pE
C pE/2
D 2pE
Correct Answer
Option A
Solution

Work done in deflecting a dipole through an angle θ\theta is given by

W=∫0θpEsin⁡θdθ=pE(1−cos⁡θ)W = \int\limits_0^\theta {pE\sin \theta d\theta } = pE\left( {1 - \cos \theta } \right)

Since θ\theta = 90° ∴\therefore W = pE(1 – cos90°) ⇒\Rightarrow W = pE

Q60
As per the diagram a point charge +q is placed at the origin O. Work done in taking another point charge −-Q from the point A [coordinates (0, aa)] to another point B
A zero
B (qQ4πε01a2).2a\left( {{{qQ} \over {4\pi {\varepsilon _0}}}{1 \over {{a^2}}}} \right).\sqrt 2 a
C (−qQ4πε01a2).2a\left( {{{ - qQ} \over {4\pi {\varepsilon _0}}}{1 \over {{a^2}}}} \right).\sqrt 2 a
D (qQ4πε01a2).a2\left( {{{qQ} \over {4\pi {\varepsilon _0}}}{1 \over {{a^2}}}} \right).{a \over {\sqrt 2 }}
Correct Answer
Option A
Solution

Work done is equal to zero because the potential of A and B are the same =

14πε0qa{1 \over {4\pi {\varepsilon _0}}}{q \over a}

No work is done if a particle does not change its potential energy. i.e. initial potential energy = final potential energy.

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