InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The small identical conducting spheres have charges of 2.0 × 10−9 C and − 0.5 × 10−9 respectively. When they are placed 4 cm apart, what is the force between them ? If they are brought into contact and then separated by 4 cm, what is the force between them ? |
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Answer» F = 9 × 10−9 Q1 Q2/d2 = 9 × 10−9 × (− 0.5 × 10−9 )/0.042 = − 56.25 × 10−7 N. When two identical spheres are brought into contact with each other and then separated, each gets half of the total charge. Hence, Q1 = Q2 = [2 × 10−9 + (− 0.5 × 10−9)/2] = 0.75 × 10−9C When they are separated by 4cm, F = 9 × 10−9 × (0.75 × 10−9)2/0.042 = 0.316 × 10−5N |
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| 2. |
If D is the electric flux density, then value of electric intensity in air is (a) D/ε0 (b) D/ε0εr(c) dV/dt (d) Q/εA |
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Answer» Correct option (a) D/ε0 |
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| 3. |
For any medium, electric flux density D is related to electric intensity E by the equation (a) D = ε0E (b) D = ε0εrE (c) D = E/ε0εr (d) D = ε0E/εr |
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Answer» Correct option (b) D = ε0εrE |
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| 4. |
Inside a conducting sphere,...remains constant (a) electric flux (b) electric intensity (c) charge (d) potential |
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Answer» Correct option (d) potential |
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| 5. |
According to Gauss’s theorem, the surface integral of the normal component of electric flux density D over a closed surface containing charge Q is (a) Q (b) Q/ε0 (c) ε0Q (d) Q2/ε0 |
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Answer» Correct option (a) Q |
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| 6. |
Capacitance (in F) of a spherical conductor with radius 1 m is(a) 1.1 × 10–10 (b) 10–6(c) 9 × 10–9 (d) 10–3 . |
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Answer» The correct option is (a) 1.1 x 10-10. Explanation: Capacitance of a spherical conductor is given by C = 4πεor where r = radius of spherical conductor C = 4πεo = 1/( 9 x 109) = 1.1 x 10-10 F |
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| 7. |
Electric flux emanating from an electric charge of + Q coulomb is (a) Q/ε0 (b) Q/εr (c) Q/ε0εr (d) Q |
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Answer» Correct option (d) Q |
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| 8. |
Two similar electric charges of 1 C each are placed 1 m apart in air. Force of repulsion between them would be nearly...... newton (a) 1 (b) 9 × 109 (c) 4 π (d) 8.854 × 10−12 |
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Answer» Correct option (b) 9 × 109 |
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| 9. |
A parallel-plate capacitor consists of two square metal plates 500 mm on a side separated by 10 mm. A slab of Teflon (εr = 2.0) 6 mm thick is placed on the lower plate leaving an air gap 4 mm thick between it and the upper plate. If 100 V is applied across the capacitor, find the electric field (E0) in the air, electric field Et in Teflon, flux density Da in air, flux density Dt in Teflon and potential difference Vt across Teflon slab. |
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Answer» C = ε0A/(d1/εe1 + d2/εr2) = (8.854 x 10-12 x (0.5)2)/((6 x 10-3/2) + (4 x 10-3/1)) = 3.16 x 10-10F = CV = 3.16 × 10−10 × 100 = 31.6 × 10−9 C D = Q/A = 31.6 × 10−9/(0.5)2 = 1.265 × 10−7 C/m2 The charge or flux density will be the same in both media i.e. Da = Dt = D In air, E0 = D/ε0 = 1.265 × 10−7/8.854 × 10−12 = 14,280 V/m In Teflon, Et = D/ε0εr = 14,280/2 = 7,140V/m Vt = Et × dt = 7,140 × 6 × 10−3 = 42.8 V |
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| 10. |
The unit of absolute permittivity of a medium is (a) joule/coulomb (b) newton-metre (c) farad/metere (d) farad/coulomb |
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Answer» Correct option (c) farad/metere |
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| 11. |
The SI unit of electric intensity is (a) N/m (b) V/m (c) N/C (d) either (b) or (c) |
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Answer» Correct option (d) either (b) or (c) |
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| 12. |
In practice, earth is chosen as a place of zero electric potential because it (a) is non-conducting (b) is easily available (c) keeps lossing and gaining electric charge every day (d) has almost constant potential. |
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Answer» Correct option (d) has almost constant potential. |
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| 13. |
A capacitor is composed of 2 plates separated by a sheet of insulating material 3 mm thick and of relative permitivity 4. The distance between the plates is increased to allow the insertion of a second sheet of 5 mm thick and of relative permitivity εr . If the equivalent capacitance is one third of the original capacitance. Find the value of εr. |
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Answer» C1(ε0 εrA/d) = k(4/3), where k = ε0A x 10+3 The composite capacitor [with one dielectric of εr1 = 4 and other dielectric of εr2 as relative permitivity has a capacitance of C/3. Two capacitors are effectively in series. Let the second dielectric contribute a capacitor of C2. K.(4/9) = C1C2/(C1 + C2) = (K.(4/3).C2)/(K.(4/3) + C2 This gives C2 = 2/3 K (2/3)K = (ε0εr2A)/(5 x 10-3) Er2 = 10/3.k 1/e0A x 10-3 = 10/3 e0 A × 103 1/e0 A × 10−3 = 10/3 = 3.33 |
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| 14. |
A point charge of 10−9 C is placed at a point A in free space. Calculate : (i) the intensity of electrostatic field on the surface of sphere of radius 5 cm and centre A.(ii) the difference of potential between two points 20 cm and 10 cm away from the charge at A. |
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Answer» (i) E = Q/4πε0 r2 = 10−9/4π × 8.854 × 10−12 × (5 × 10−2)2 = 3,595 V/m (ii) Potential of first point = Q/4πε0d = 10−9/4π × 8.854 × 10−12 × 0.2 = 45V Potential of second point = 10−9 /4π × 8.854 × 10−12 × 0.1 = 90V ∴ p.d. between two points = 90 − 45 = 45 V |
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