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NEET • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential and capacitance practice · Medium

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Showing 31-60 of 175 questions
31 Medium
31 of 175
Understanding +4 | -1
An electric dipole of moment p placed in a uniform electric field E has minimum potential energy when the angle between p and E is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the potential energy of a dipole in an electric field is minimum when the dipole moment is aligned with the electric field (angle is zero).
32 Medium
32 of 175
Understanding +4 | -1
Can a metal be used as a medium for dielectric
View answer and explanation
Correct option: B
No
Option B (No) is correct because metals are conductors, and a dielectric material must be an insulator.
33 Medium
33 of 175
Application +4 | -1
An electric dipole is placed making at an angle 60 ∘C60\,^\circ C with an electric field of strength 4×105  N/C4 \times {10^5}\,\,N/C. It experiences a torque equal to 83  N−m8\sqrt 3 \,\,N - m. Calculate the charge on the dipole, if it is of length 4 cm.
View answer and explanation
Correct option: B
10−3C{10^{ - 3}}C
Option B (10−3C{10^{ - 3}}C) is correct because the charge is calculated using the formula for torque τ=pEsin⁡θ=qdEsin⁡θ\tau = pE \sin \theta = qdE \sin \theta, where θ=60°\theta = 60°.
34 Medium
34 of 175
Application +4 | -1
A molecule with a dipole moment pp is placed in an electric field of strength E. Initially the dipole is aligned parallel to the field. If the dipole is to be related to be anti-parallel to the field the work required to be done by an external agency is
View answer and explanation
Correct option: D
2pE2pE
Option D (2pE2pE) is correct because the work required to rotate the dipole from parallel to anti-parallel is equal to the change in potential energy, which is 2pE2pE.
35 Medium
35 of 175
Application +4 | -1
There is an air filled 1pF1pF parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to 2pF2pF. The dielectric constant of wax is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the dielectric constant is calculated using the formula for capacitance with a dielectric, considering the change in plate separation.
36 Medium
36 of 175
Analyzing +4 | -1
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
View answer and explanation
Correct option: C
Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
Option C (Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates) is correct because inserting a dielectric increases capacitance, which decreases the potential difference and stored energy while charge remains constant.
37 Medium
37 of 175
Understanding +4 | -1
Ten electrons are equally spaced and fixed around a circle of radius R.R. Relative to V=0V = 0 at infinity, the electrostatic potential VV and the electric field EE at the centre CC are
View answer and explanation
Correct option: B
V≠0  and  E⃗=0V \ne 0\;\mathrm{and\;}\vec E = 0
Option B (V≠0  and  E⃗=0V \ne 0\;\mathrm{and\;}\vec E = 0) is correct because the potential due to multiple charges is non-zero, but the electric field at the center due to symmetrically placed equal charges is zero.
38 Medium
38 of 175
Analyzing +4 | -1
Statement-I: In bringing an electron towards a proton electrostatic potential energy of the system increases.

Statement-II: Potential due to proton is positive.
View answer and explanation
Correct option: D
Statement-I is false, Statement-II is false
Option C (Statement-I is false, Statement-II is true) is incorrect. Statement-I is false because bringing an electron towards a proton decreases the potential energy of the system. Statement-II is true because protons have a positive charge, and thus create a positive potential.
39 Medium
39 of 175
Application +4 | -1
A non-conducting ring of radius 0.5  m0.5\;m carries a total charge of 1.11×10−10C1.11 \times {10^{ - 10}}C distributed non-uniformly on its circumference producing an electric field E⃗\vec E everywhere in space. The value of the line integral ∫l=0l=∞−E⃗.dl→  (l=0\int_{l = 0}^{l = \infty } { - \vec E.\overrightarrow {dl} \;(l = 0} being centre of the ring) in volt is
View answer and explanation
Correct option: A
+2+ 2
Option A (+2+ 2) is correct because the line integral of the electric field from the center to infinity is equal to the potential difference between those points, which is the potential at the center.
40 Medium
40 of 175
Understanding +4 | -1
When one electron is taken towards the other electron, then the electric potential energy of the system
View answer and explanation
Correct option: B
Increases
Option B (Increases) is correct because bringing two electrons closer increases the potential energy due to repulsive forces.
41 Medium
41 of 175
Application +4 | -1
A cylindrical capacitor has charge QQ and length L.L. If both the charge and length of the capacitor are doubled, by keeping other parameters fixed, the energy stored in the capacitor
View answer and explanation
Correct option: B
Increases two times
Option B (Increases two times) is correct because doubling both charge and length doubles the energy stored in the cylindrical capacitor.
42 Medium
42 of 175
Understanding +4 | -1
A parallel plate capacitor is charged. If the plates are pulled apart
View answer and explanation
Correct option: B
The potential difference increases
Option B (The potential difference increases) is correct because pulling the plates apart decreases the capacitance, and since the charge remains constant, the potential difference must increase.
43 Medium
43 of 175
Remembering +4 | -1
In which form energy is stored in a capacitor:
View answer and explanation
Correct option: A
Electric field
Option A (Electric field) is correct because the energy in a capacitor is stored in the form of an electric field between its plates.
44 Medium
44 of 175
Application +4 | -1
A sample of HClHCl gas is placed in an electric field of 3×104  NC−1.3 \times {10^4}\;N{C^{ - 1}}. The dipole moment of each HClHCl molecule is 6×10−30c×m.6 \times {10^{ - 30}}c \times m. The maximum torque that can act on a molecule is
View answer and explanation
Correct option: C
18×10−26  Nm18 \times {10^{ - 26}}\;Nm
Option C (18×10−26  Nm18 \times {10^{ - 26}}\;Nm) is correct because it is the maximum torque calculated using the formula torque = dipole moment * electric field.
45 Medium
45 of 175
Application +4 | -1
The electric potential VV at any point (x,y,z)\left( {x,y,z} \right), all in meters in space is given by V=4x2  volt.V = 4{x^2}\;volt. The electric field at the point (1,0,2)\left( {1,0,2} \right) in volt/metervolt/meter, is
View answer and explanation
Correct option: B
8 along negative XX-axis
Option B (8 along negative XX-axis) is correct because the electric field is the negative gradient of the potential, and in this case, it points in the negative x-direction.
46 Medium
46 of 175
Understanding +4 | -1
The electric potential due to a small electric dipole at a large distance rr from the centre of the dipole is proportional to
View answer and explanation
Correct option: B
1/r1/r
Option B (1/r1/r) is correct because the electric potential due to a dipole decreases as the inverse of the distance from the dipole.
47 Medium
47 of 175
Analyzing +4 | -1
Assertion:- If three capacitors of capacitances C1<C2<C3\mathrm{C}_1 < \mathrm{C}_2 < \mathrm{C}_3 are connected in series and parallel, then Cparallel >Cseries \mathrm{C}_{\text {parallel }} > \mathrm{C}_{\text {series }}.
Reason:- Cseries =C1+C2+C3\mathrm{C}_{\text {series }}=\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 and 1Cparallel =1C1+1C2+1C3\frac{1}{\mathrm{C}_{\text {parallel }}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because the formulas for series and parallel combinations are stated incorrectly in the Reason.
48 Medium
48 of 175
Understanding +4 | -1
When air in a capacitor is replaced by a medium of dielectric constant K,K, the capacity
View answer and explanation
Correct option: B
Increases KK times
Option B (Increases KK times) is correct because the capacitance of a capacitor increases by a factor of K when a dielectric material with dielectric constant K is inserted between the plates.
49 Medium
49 of 175
Application +4 | -1
A parallel plate capacitor has circular plates of 0.08m radius and 1.0×10−3m1.0 \times {10^{ - 3}}m separation. If a P.D. of 100  volt100\;volt is applied, the charge will be
View answer and explanation
Correct option: B
1.8×10−8C1.8 \times {10^{ - 8}}C
Option B (1.8×10−8C1.8 \times {10^{ - 8}}C) is correct because it is the charge calculated using the capacitance and potential difference.
50 Medium
50 of 175
Remembering +4 | -1
Let VV be the electric potential at a given point. Then the electric field Ex{E_x} along xx-direction at that point is given by
View answer and explanation
Correct option: C
−dVdx- \frac{{dV}}{{dx}}
Option C (−dVdx- \frac{{dV}}{{dx}}) is correct because the electric field is the negative gradient of the electric potential.
51 Medium
51 of 175
Analyzing +4 | -1
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason
R.
Assertion A : Two metallic spheres are charged to the same potential. One of them is hollow and
another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one.
Reason R: Capacitance of metallic spheres depend on the radii of spheres.
In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: B
A is false but R is true
Option B is correct because Assertion A is false (solid sphere will have the same charge as the hollow one for the same potential), but Reason R is true (capacitance depends on radii).
52 Medium
52 of 175
Remembering +4 | -1
The energy stored in the condenser is
View answer and explanation
Correct option: B
12QV\frac{1}{2}QV
Option B (12QV\frac{1}{2}QV) is correct because it represents the formula for the energy stored in a capacitor, derived from integrating the work done to charge the capacitor.
53 Medium
53 of 175
Understanding +4 | -1
The electric potential at a point on the axis of an electric dipole depends on the distance rr of the point from the dipole as
View answer and explanation
Correct option: B
∝1r2\propto \frac{1}{{{r^2}}}
Option B (∝1r2\propto \frac{1}{{{r^2}}}) is correct because the electric potential at a point on the axis of an electric dipole is inversely proportional to the square of the distance from the dipole.
54 Medium
54 of 175
Remembering +4 | -1
The condensers of capacity C1and  C2 {C_1}\mathrm{ and }\,\,{C_2}\, are connected in parallel, then the equivalent capacitance is:
View answer and explanation
Correct option: A
C1+C2{C_1} + {C_2}
Option A (C1+C2{C_1} + {C_2}) is correct because the equivalent capacitance of capacitors in parallel is the sum of their individual capacitances.
55 Medium
55 of 175
Application +4 | -1
Two opposite and equal charges 4×10−8coulomb4 \times {10^{ - 8}}coulomb when placed 2×10−2cm2 \times {10^{ - 2}}cm away, form a dipole. If this dipole is placed in an external electric field 4×108newton/coulomb,4 \times {10^8}newton/coulomb, the value of maximum torque and the work done in rotating in through 180o{180^\mathrm{o}} will be
View answer and explanation
Correct option: D
32×10−4Nm32 \times {10^{ - 4}}Nm and 64×10−4J64 \times {10^{ - 4}}J
Option D (32×10−4Nm32 \times {10^{ - 4}}Nm and 64×10−4J64 \times {10^{ - 4}}J) is correct because it correctly calculates the maximum torque and work done using the dipole moment and electric field.
56 Medium
56 of 175
Application +4 | -1
In a Vande graff generator a spherical metal shell is to be a 15×105 V15 \times {10^5}\,V electrode. The dielectric strength of a gas surrounding the electrodes is 5×107 Vm−15 \times {10^7}\,V{m^{ - 1}}. The minimum radius of the spherical shell is
View answer and explanation
Correct option: C
3 cm
Option C (3 cm) is correct because the electric field at the surface of the sphere must be less than or equal to the dielectric strength of the surrounding gas.
57 Medium
57 of 175
Understanding +4 | -1
An electric dipole has the magnitude of its charge as qq and its dipole moment is p.p. It is placed in a uniform electric field E.E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively
View answer and explanation
Correct option: C
Zero and minimum
Option C (Zero and minimum) is correct because when the dipole moment is aligned with the electric field, the net force is zero and the potential energy is at its minimum.
58 Medium
58 of 175
Understanding +4 | -1
Equal charges are given to two spheres of different radii. The potential will
View answer and explanation
Correct option: A
Be more on the smaller sphere
Option A (Be more on the smaller sphere) is correct because the electric potential is inversely proportional to the radius of the sphere for a given charge.
59 Medium
59 of 175
Analyzing +4 | -1
Assertion: The electric field due to a dipole on its axis line at a distance rr is E.E. Then, electric field due to the same dipole on the equatorial line and at the same distance will be E/2.E/2.
Reason: Electric field due to dipole varies inversely as the square of the distance.
View answer and explanation
Correct option: C
Assertion is true but reason is false
Option C (Assertion is true but reason is false) is correct because the electric field due to a dipole varies inversely as the cube of the distance, not the square.
60 Medium
60 of 175
Application +4 | -1
A parallel plate capacitor has plate area AA and separation d.d. It is charged to a potential difference V0.{V_0}. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is
View answer and explanation
Correct option: D
ε0AV02d\frac{{{\varepsilon _0}AV_0^2}}{d}
Option D (ε0AV02d\frac{{{\varepsilon _0}AV_0^2}}{d}) is correct because it represents the work done to increase the separation, calculated from the change in energy stored in the capacitor.
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