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Electrostatics

NEET Physics · 98 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
Two identical charged spheres suspended from a common point by two massless strings of lengths ll, are initially at a distance d(d < < ll) 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
A v ∝ \propto x −-1/2
B v ∝ \propto x −-1
C v ∝ \propto x 1/2
D v ∝ \propto x
Correct Answer
Option A
Solution

From figure

tan⁡θ=Femg≃θ\tan \theta = {{{F_e}} \over {mg}} \simeq \theta
kq2x2mg=x2ℓ{{k{q^2}} \over {{x^2}mg}} = {x \over {2\ell }}

⇒\Rightarrow

x3∝q2{x^3} \propto {q^2}

...(i)

⇒x3/2∝q\Rightarrow {x^{3/2}} \propto q

Differentiating eq. (1) w.r.t. time

3x2dxdt∝2qdqdt3{x^2}{{dx} \over {dt}} \propto 2q{{dq} \over {dt}}

but

dqdt{{dq} \over {dt}}

is constant so x 2 (v) ∝\propto q Replace q from eq. (2) x 2 (v) ∝\propto x 3/2 ⇒\Rightarrow v ∝\propto x –1/2

Q32
If potential (in volts) in a region is expressed as V(x, y, z) = 6xy −- y + 2yz, the electric field (in N/C) at point (1, 1, 0) is
A −(2i^+3j^+k^) - \left( {2\widehat i + 3\widehat j + \widehat k} \right)
B −(6i^+9j^+k^) - \left( {6\widehat i + 9\widehat j + \widehat k} \right)
C −(3i^+5j^+3k^) - \left( {3\widehat i + 5\widehat j + 3\widehat k} \right)
D −(6i^+5j^+2k^) - \left( {6\widehat i + 5\widehat j + 2\widehat k} \right)
Correct Answer
Option D
Solution

Potential in a region V = 6xy – y + 2yz As we know the relation between electric potential and electric field is

E→=−dVdx\overrightarrow E = {{ - dV} \over {dx}}
E→=(∂V∂xi^+∂V∂yj^+∂V∂zk^)\overrightarrow E = \left( {{{\partial V} \over {\partial x}}\widehat i + {{\partial V} \over {\partial y}}\widehat j + {{\partial V} \over {\partial z}}\widehat k} \right)
E→=[(6yi^+(6x−1+2z)j^+(2y)k^)]\overrightarrow E = \left[ {\left( {6y\widehat i + \left( {6x - 1 + 2z} \right)\widehat j + \left( {2y} \right)\widehat k} \right)} \right]
E→(1,1,0)=−(6i^+5j^+2k^){\overrightarrow E _{\left( {1,1,0} \right)}} = - \left( {6\widehat i + 5\widehat j + 2\widehat k} \right)
Q33
The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by
A 4πε0Aa34\pi {\varepsilon _0}A{a^3}
B ε0Aa3{\varepsilon _0}A{a^3}
C 4πε0Aa24\pi {\varepsilon _0}A{a^2}
D Aε0a2A{\varepsilon _0}{a^2}
Correct Answer
Option A
Solution

According to question, electric field varies as E = Ar Here r is the radial distance.

At r = a, E = Aa …(i) Net flux emitted from a spherical surface of radius a is

ϕnet=qenε0{\phi _{net}} = {{{q_{en}}} \over {{\varepsilon _0}}}
⇒(Aa)×(4πa2)=qε0\Rightarrow \left( {Aa} \right) \times \left( {4\pi {a^2}} \right) = {q \over {{\varepsilon _0}}}

[Using equation (i)]

∴q=4πε0Aa3\therefore q = 4\pi {\varepsilon _0}A{a^3}
Q34
A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere rrespectively are
A zero and Q4πε0R2{Q \over {4\pi {\varepsilon _0}{R^2}}}
B Q4πε0R{Q \over {4\pi {\varepsilon _0}R}} and zero
C Q4πε0R{Q \over {4\pi {\varepsilon _0}R}} and Q4πε0R2{Q \over {4\pi {\varepsilon _0}{R^2}}}
D both are zero
Correct Answer
Option B
Solution

For the conducting sphere, Potential at the centre = Potential on the sphere

=14πε0QR= {1 \over {4\pi {\varepsilon _0}}}{Q \over R}

Electric field at the centre = 0

Q35
In a region, the potential is represented by V(x, y, z) = 6x −- 8xy −- 8y + 6yz, where VV is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is
A 656\sqrt 5 N
B 30 N
C 24 N
D 4354\sqrt {35} N
Correct Answer
Option D
Solution
E→=−∂V∂xi^−∂V∂yj^−∂V∂zk^\overrightarrow E = - {{\partial V} \over {\partial x}}\widehat i - {{\partial V} \over {\partial y}}\widehat j - {{\partial V} \over {\partial z}}\widehat k
=−[(6−8y)i^+(−8x−8+6z)j^+(6y)k^]= - \left[ {\left( {6 - 8y} \right)\widehat i + \left( { - 8x - 8 + 6z} \right)\widehat j + \left( {6y} \right)\widehat k} \right]

At (1, 1, 1),

E→=2i^+10j^−6k^\overrightarrow E = 2\widehat i + 10\widehat j - 6\widehat k
⇒(E→)=22+102+62=140=235\Rightarrow \left( {\overrightarrow E } \right) = \sqrt {{2^2} + {{10}^2} + {6^2}} = \sqrt {140} = 2\sqrt {35}
∴F=qE→=2×235=435\therefore F = q\overrightarrow E = 2 \times 2\sqrt {35} = 4\sqrt {35}
Q36
A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to
A −-Q/4
B Q/4
C −-Q/2
D Q/2
Correct Answer
Option A
Solution

Let the distance between given changes be 2x In equilibrium,

∣F→AB∣+∣F→AC∣=0\left| {{{\overrightarrow F }_{AB}}} \right| + \left| {{{\overrightarrow F }_{AC}}} \right| = 0

So,

∣FAB∣=−∣FAC∣\left| {{F_{AB}}} \right| = - \left| {{F_{AC}}} \right|
14πε0Q24x2=−14πε0Qqx2{1 \over {4\pi {\varepsilon _0}}}{{{Q^2}} \over {4{x^2}}} = - {1 \over {4\pi {\varepsilon _0}}}{{Qq} \over {{x^2}}}
⇒q=−Q/4\Rightarrow q = - Q/4
Q37
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 o is
A p 2 E
B pE
C infinity
D pE 2
Correct Answer
Option B
Solution

Potential energy of dipole,

U=−pE(cos⁡θ2−cos⁡θ1)U = - pE\left( {\cos {\theta _2} - \cos {\theta _1}} \right)

Here,

θ1=0o,θ2=90o{\theta _1} = {0^o},{\theta _2} = {90^o}

∴\therefore U = – pE(cos90° – cos0°) = – pE(0 – 1) = pE

Q38
Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become
A (2r3)\left( {{{2r} \over {\sqrt 3 }}} \right)
B (2r3)\left( {{{2r} \over 3}} \right)
C (12)2{\left( {{1 \over {\sqrt 2 }}} \right)^2}
D (r23)\left( {{r \over {\sqrt[3]2 }}} \right)
Correct Answer
Option D
Solution

From figure,

tan⁡θ=Femg⇒r/2y=kq2r2mg\tan \theta = {{{F_e}} \over {mg}} \Rightarrow {{r/2} \over y} = {{{{k{q^2}} \over {{r^2}}}} \over {mg}}
[∵F=kq2r2from coulomb’s law]\left[\because {F = {{k{q^2}} \over {{r^2}}}} { from \,coulomb’s \,law} \right]
⇒r3∝y⇒r′3∝y2⇒r′r=121/3\Rightarrow {r^3} \propto y \Rightarrow r{'^3} \propto {y \over 2} \Rightarrow {{r'} \over r} = {1 \over {{2^{1/3}}}}
⇒r′=r23\Rightarrow r' = {r \over {\sqrt[3]2 }}
Q39
A, B and C are three points in a uniform electric field. The electric potential is
A maximum at C
B same at all the three points A, B and C
C maximum at A
D maximum at B
Correct Answer
Option D
Solution

In the direction of electric field, electric potential decreases. ∴\therefore V B > V C > V A

Q40
Two metallic spheres of radii 1 cm and 3 cm are given charges of −-1 × \times 10 −-2 C and 5 × \times 10 −-2 C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
A 2 × \times 10 −-2 C
B 3 × \times 10 −-2 C
C 4 × \times 10 −-2 C
D 1 × \times 10 −-2 C
Correct Answer
Option B
Solution

At equilibrium potential of both sphere becomes same if charge of sphere one x and other sphere Q – x then where Q = 4 × 10 –2 C v 1 = v 2

kx1 cm=k(Q−x)3 cm{{kx} \over {1\,cm}} = {{k\left( {Q - x} \right)} \over {3\,cm}}

3x = Q – x ⇒\Rightarrow 4x = Q

x=Q4=4×10−24C=1×10−2x = {Q \over 4} = {{4 \times {{10}^{ - 2}}} \over 4}C = 1 \times {10^{ - 2}}

Q' = Q – x = 3 × 10 –2 C

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