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Electrostatics

NEET Physics · 98 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
In a certain region of space with volume 0.2 m 3 , the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is :
A 0.5 N/C
B 1 N/C
C 5 N/C
D zero
Correct Answer
Option D
Solution

Potential V is constant.

Electric field E is the differentiation of potential V with respect to distance.

Differentiation of a constant is zero.

So E = 0.

Q22
A spherical conductor of radius 10 cm has a charge of 3.2 × \times 10 -7 C distributed uniformly. That is the magnetude of electric field at a point 15 cm from the centre of the sphere? (14πε0=9×109Nm2/c2)\left( {{1 \over {4\pi {\varepsilon _0}}} = 9 \times {{10}^9}N{m^2}/{c^2}} \right)
A 1.28 × \times 10 5 N/C
B 1.28 × \times 10 6 N/C
C 1.28 × \times 10 7 N/C
D 1.28 × \times 10 4 N/C
Correct Answer
Option A
Solution

The electric field for the conducting sphere, away from the surface is given by,

E=14πε0Qr2E = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{r^2}}}
=9×109×3.2×10−7225×10−4= {{9 \times {{10}^9} \times 3.2 \times {{10}^{ - 7}}} \over {225 \times {{10}^{ - 4}}}}
=0.128×106= 0.128 \times {10^6}
=1.28×105= 1.28 \times {10^5}

N/C

Q23
A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre :
A zero as r increases for r < R, decreases as r increases for r > R
B zero as r increases for r < R, increases as r increases for r > R.
C decreases as r increases for r < R and for r > R.
D increases as r increases for r < R and for r > R.
Correct Answer
Option A
Solution

For inside (r << R) Apply, Gauss law,

∮Ein.dS→\oint {{E_{in}}} .\overrightarrow {dS}

=

qenε0{{{q_{en}}} \over {{\varepsilon _0}}}

⇒\Rightarrow E in = 0 [ As q en = 0 ] For outside (r >> R)

∮E→out.dS→=qenε0\oint {{{\overrightarrow E }_{out}}} .\overrightarrow {dS} = {{{q_{en}}} \over {{\varepsilon _0}}}

Here, q en = Q So, E out .4π\pir 2 =

Qε0{Q \over {{\varepsilon _0}}}

∴\therefore E out ∝\propto

1r2{1 \over {{r^2}}}
Q24
Two point charges A and B, having charges +Q and – Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes :
A F
B 16F9{{16F} \over 9}
C 9F16{{9F} \over {16}}
D 4F3{{4F} \over 3}
Correct Answer
Option C
Solution

Initially F =

−KQ2r2- {{K{Q^2}} \over {{r^2}}}

If 25% of charge of A transferred to B then after the transformation of charge. q A =

Q−Q4Q - {Q \over 4}

=

3Q4{{3Q} \over 4}

and q b = -Q +

Q4{Q \over 4}

=

−3Q4- {{3Q} \over 4}

New force, F 1 =

K(3Q4)(−3Q4)r2{{K\left( {{{3Q} \over 4}} \right)\left( { - {{3Q} \over 4}} \right)} \over {{r^2}}}
=−916KQ2r2= - {9 \over {16}}{{K{Q^2}} \over {{r^2}}}

=

9F16{{9F} \over {16}}
Q25
Two parallel infinite line charges with linear charge densities +λ\lambda C/m and -λ\lambda C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?
A 2λπε0R{{2\lambda } \over {\pi {\varepsilon _0}R}} N/C
B zero
C λπε0R{\lambda \over {\pi {\varepsilon _0}R}} N/C
D λ2πε0R{\lambda \over {2\pi {\varepsilon _0}R}} N/C
Correct Answer
Option C
Solution
E→net=E→1+E→2{\overrightarrow E _{net}} = {\overrightarrow E _1} + {\overrightarrow E _2}

=

λ2πε0R{\lambda \over {2\pi {\varepsilon _0}R}}

+

λ2πε0R{\lambda \over {2\pi {\varepsilon _0}R}}

=

λπε0R{\lambda \over {\pi {\varepsilon _0}R}}

N/C

Q26
A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E→\overrightarrow E . Due to the force qE→\overrightarrow E , its velocity increases from 0 to 6 m s –1 in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively
A 2 m s –1 , 4 m s –1
B 1 m s –1 , 3 m s –1
C 1 m s –1 , 3.5 m s –1
D 1.5 m s –1 , 3 m s –1
Correct Answer
Option B
Solution

Acceleration, a =

v−ut{{v - u} \over t}

=

6−01{{6 - 0} \over 1}

= 6 ms -2 For t = 0 to t = 1 s, S 1 =

12×6(1)2{1 \over 2} \times 6{\left( 1 \right)^2}

= 3 m ...(i) For t = 1 s = to t = 2 s, S 2 = 6.1 -

12×6(1)2{1 \over 2} \times 6{\left( 1 \right)^2}

= 3 m ...(ii) For t = 2 s to t = 3 s, S 3 = 0 -

12×6(1)2{1 \over 2} \times 6{\left( 1 \right)^2}

= -3 m ...(ii) Total displacement S = S 1 + S 2 + S 3 = 3 m Average velocity =

33{3 \over 3}

= 1 ms -1 Total distance travelled = 9 m Average speed =

93{9 \over 3}

= 3 ms -1

Q27
An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is
A smaller
B 5 times greater
C 10 times greater
D equal
Correct Answer
Option A
Solution

Force experienced by a charged particle in an electric field, F = qE As F = ma ∴\therefore ma = qE ⇒\Rightarrow a =

qEm{{qE} \over m}

As electron and proton both fall from same height at rest. Then initial velocity = 0 From the formula, s = ut +

12at2{1 \over 2}a{t^2}

∴\therefore h =

12at2{1 \over 2}a{t^2}

⇒\Rightarrow h =

12qEmt2{1 \over 2}{{qE} \over m}{t^2}

∴\therefore t =

2hmqE\sqrt {{{2hm} \over {qE}}}

⇒\Rightarrow t ∝\propto

m\sqrt m

as ‘q’ is same for electron and proton. Since, electron has smaller mass so it will take smaller time.

Q28
The diagrams below show regions of equipotentials. A positive charge is moved from A to B in each diagram.
A In all the four cases the work done is the same.
B Minimum work is required to move q in figure(I).
C Maximum work is required to move q in figure (II).
D Maximum work is required to move q in figure (III)
Correct Answer
Option A
Solution

As the regions are of equipotentials, so Work done W = q

Δ\Delta

V

Δ\Delta

V is same in all the cases hence work - done will also be same in all the cases.

Q29
Suppose the charge of a proton and an electron differ slightly. One of them is −-e, the other is (e + Δ\Delta e). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (musch greater than atomic size) apart is zero, then Δ\Delta e is of the order of [Given : mass of hydrogen m h = 1.67 × \times 10−- 27 kg]
A 10 −-23 C
B 10 −-37 C
C 10 −-47 C
D 10 −-20 C
Correct Answer
Option B
Solution

According to question, the net electrostatic force (F E ) = gravitational force (F G ) F E = F G

⇒14πε0Δe2d2=Gm2d2\Rightarrow {1 \over {4\pi {\varepsilon _0}}}{{\Delta {e^2}} \over {{d^2}}} = {{G{m^2}} \over {{d^2}}}
⇒Δe=mGK(14πε0=k=9×109)\Rightarrow \Delta e = m\sqrt {{G \over K}} \left( {{1 \over {4\pi {\varepsilon _0}}} = k = 9 \times {{10}^9}} \right)
=1.67×10−276.67×10−119×109= 1.67 \times {10^{ - 27}}\sqrt {{{6.67 \times {{10}^{ - 11}}} \over {9 \times {{10}^9}}}}
Δe≈1.436×10−37C\Delta e \approx 1.436 \times {10^{ - 37}}C
Q30
An electric dipole is placed at an angle of 30 o with an electric field intensity 2 × \times 10 5 N C −-1 . It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is
A 8 mC
B 2 mC
C 5 mC
D 7 μ\mu C
Correct Answer
Option B
Solution

Torque is given by τ = pE sin θ τ = pE sin θ = q

ll

E sinθ ⇒\Rightarrow q = τ /

ll

E sinθ = 4/(2 × 10 -2 × 0.5 × 2 × 10 5 ) = 2 mC

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