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Correct choice is (c) Zero

To explain I would say: Net ELECTROSTATIC field is zero in the interior of a CONDUCTOR. When a conductor is PLACED in an electric field, its free electrons begin to move in the opposite direction. Negative charges are induced on the left end and positive charges on the right end of the conductor. The process continues until the electric field set up by induced charges BECOMES EQUAL and opposite the external field.

Correct choice is (b) Electric SUSCEPTIBILITY

The EXPLANATION is: The ratio of the polarization to εo times the electric FIELD is CALLED the electric susceptibility of the dielectric. It DESCRIBES the electrical behaviour of a dielectric.

Correct answer is (c) \(\FRAC {(q \, LOG2)}{a}\)

Best explanation: The total potential energy at the origin DUE to the system is V=\(\frac {q}{a} – \frac {q}{2a} + \frac {q}{3a} – \frac {q}{4a} + \frac {q}{5a}\)……(up to infinity), where a, 2a, 3a… are the distances of the charges from the origin. Now, taking \(\frac {q}{a}\) common from the terms, we get V=\(\frac {q}{a} (1 – \frac {1}{2} + \frac {1}{3} – \frac {1}{4} + \frac {1}{5})\). By substituting the expansion of log2, we can get V=\(\frac {(q \, log2)}{a}\).

Right answer is (a) PLANE SURFACE

To explain: If a point charge is situated at infinity, the ELECTRIC field lines coming out of it will be in the form of parallel straight lines. As we know that field lines cut the equipotential surfaces orthogonally, THEREFORE the equipotential surfaces must be plane surfaces. They can be considered the surface of a sphere of infinite RADIUS.

Correct choice is (a) Electric potential

Easy explanation: Electric potential is DEFINED as the amount of work DONE to bring a unit positive charge from an infinite distance to a particular point of an electric field. The total ENERGY of that point charge MEANS the sum of KINETIC energy and potential energy which is not the same as the potential energy if the particle is in motion.

The CORRECT CHOICE is (a) A Van de Graff generator PRODUCES large voltage and LESS current

Explanation: A Van de Graff generators produces large voltage and less current. A Van de Graff generator is an electrostatic generator which creates very high electric potentials. It produces very high voltage DIRECT current electricity at low current levels.

Right answer is (d) 2.25 mJ

The explanation is: Common potential, V = \( \FRAC {(C_1V_1 + C_1V_2)}{(C_1 + C_2)}\)

V = \( \frac {[(5 \TIMES 50) + (7 \times 40)] \times 10^{-6}}{(5 + 7) \times 10^{-6}}\)

V = \( \frac {(250 + 280)}{12}\) = 44.4 V

ENERGY lost by 5 μF capacitor = \(\frac {1}{2}\)C1V1^2 – \(\frac {1}{2}\)C1V^2 = \(\frac {1}{2}\)C (V1^2 – V^2)

\(\frac {1}{2}\) × (5 × 10^-6) × (50^2 – 40^2) = 0.00225

U = 2.25 × 10^-3 J

U = 2.25 mJ

Therefore, the energy lost is 2.25 mJ

The correct option is (b) Dielectric

The explanation: A dielectric is a SUBSTANCE which does not ALLOW the flow of charges through it but permits them to EXERT electrostatic forces on one ANOTHER through it. A dielectric is essentially an insulator which can be polarized through small localized DISPLACEMENTS of its charges.

The correct choice is (d) V=\(\FRAC {3q}{16\pi \varepsilon_o}\)

Easiest explanation: We know electric potential(V) due to CHARGE ‘q’ is V = \(\frac {kq}{r}\).

Electric potential (V) is a scalar quantity, so, the total potential at X = 0 is the sum of all the individual charges

V=[\((\frac {kq}{1}) + (\frac {kq}{4}) + (\frac {kq}{16}) + (\frac {kq}{64}) \) + ……..]

V=kq\((\frac {1}{(1-\frac{1}{4})})\)→ sum of infinite G.P=\(\frac {a}{(1-r)}\)

V=\((\frac {q}{4\pi \varepsilon_o}) \times (\frac {3}{4})\)

V=\(\frac {3q}{16\pi \varepsilon_o}\)

Correct answer is (B) Cs = \(\frac {1}{C_1} + \frac {1}{C_2}\)

For explanation: When 2 capacitors are connected in SERIES, the equivalent capacitance Cs is given by

Cs = \(\frac {1}{C_1} + \frac {1}{C_2}\)

When the capacitors are connected in series, the charge PASSING through each capacitor is the same.

The correct option is (a) Concentrically

Best EXPLANATION: We know that for a conducting sphere, the charge is always distributed on its outer SURFACE. We also know that charge FLOWS from higher potential to lower potential. But if we PUT the SMALLER sphere inside the larger sphere and connect them with a conductor, they will act as a single conductor and charge will be distributed to its outer surface, i.e. charge will flow to the larger sphere.

The correct option is (d) Zero

To EXPLAIN: As the surface is EQUIPOTENTIAL, all the points on the surface of the plane have the same potential values. THEREFORE work done to move the charge by a distance x on that surface is zero. The electric FIELD plays no role here.

Correct answer is (d) 4

To elaborate: Let the distance between any TWO charges is d. Therefore the net potential energy of the SYSTEM is\(\FRAC {-Q*Q}{d}+\frac {-q*Q}{d}+\frac {-q*q}{2d}\). But the total energy of the system is zero. So, -qQ-qQ+\(\frac {q^2}{2}\)=0 that means q=4Q, i.e. \(\frac {q}{Q}\)=4.

Right option is (B) 2.65*10^6 m/s

To elaborate: Increase in POTENTIAL energy of the electron when it strikes another plate = Potential difference*charge of electron=20*1.6*10^-19 J=3.201*10^-18J and as the electron was at rest initially, this energy will be converted to KINETIC energy completely. Therefore, 0.5*MASS of electron*(velocity)^2=3.201*10^-18J. SUBSTITUTING m=9.11*10^-31kg, we get v=2.65*10^6 m/s.

Right choice is (d) 1933, Robert Van de Graff

The BEST EXPLANATION: Van de Graff GENERATOR was invented by Robert Jemison Van de Graff on November 28, 1933. Robert Jemison invented the Van de Graff generator, which is a kind of high-voltage ELECTROSTATIC generator that accelerates particles, while he was doing his PhD in Princeton University.

Correct answer is (C) Van de Graff generator is used to create a large amount of static electricity

The best explanation: Van de Graff generator is used to create a large amount of static electricity. A Van de Graff generator uses static electricity and a MOVING belt to charge a large metal sphere to a very high voltage. As the belt moves, electrons move from the rubber belt to the silicon ROLLER, causing the belt to become positively charged and the roller to become NEGATIVELY charged. As a RESULT, it builds up positive charge.

Right choice is (c) \(\frac {1}{C} = \frac {1}{C_1} + \frac {1}{C_2} + \frac {1}{C_3} + ……. + \frac {1}{C_n}\)

Best EXPLANATION: The equivalent capacitance of the parallel plate capacitors CONNECTED in series is given by the sum of the RECIPROCALS of the individual capacitances. That is MATHEMATICALLY,

\(\frac {1}{C} = \frac {1}{C_1} + \frac {1}{C_2} + \frac {1}{C_3} + ……. + \frac {1}{C_n}\)

The correct choice is (B) False

To explain: The molecules of a polar dielectric have PERMANENT dipole MOMENTS. In the absence of an external electric field, the dipole moments of different molecules are randomly oriented DUE to thermal agitation in the MATERIAL.

The correct option is (a) True

The best EXPLANATION: YES, the energy in a CAPACITOR is stored in electric fields, between its plates. A capacitor is a device that is USED to store energy or charges. It is ALSO used for extracting energy from an electric field. When a capacitor is being charged, an electric field is builds up.

The CORRECT choice is (c) When the CAPACITORS are connected in parallel

Explanation: The EFFECTIVE capacitance of a capacitor is increased when the capacitors are connected in parallel. When the capacitors are connected in parallel, the equivalent capacitance is given by:

Cp = C1 + C2 + C3 +…….

When the capacitors are connected in parallel, the POTENTIAL difference across each capacitor is the same.

Right choice is (b) 0.1V

For explanation I WOULD say: Work done to MOVE a charge q in a potential difference of V is equal to q*V. According to the QUESTION, 2J=20C*Potential difference. Therefore, potential difference=\(\frac {2}{20}\)V=0.1V. The definition of 1J thus becomes, work done to move a charge of 1C from ONE point to another where the potential difference between the two points is 1V.

Right option is (a) U = \(\FRAC {1}{(4\pi \varepsilon_o)} \times [ \frac {q_1q_2}{r} ]\)

EXPLANATION: ELECTRIC potential energy of a system of charges is the total amount of work done in BRINGING the various charges to their RESPECTIVE positions from infinitely large mutual separations.

The expression for electric potential energy is given by:

U = \(\frac {1}{(4\pi \varepsilon_o)} \times [ \frac {q_1q_2}{r} ]\)

Correct answer is (b) False

To elaborate: ELECTRIC potential depends only on the electric field intensity and the amount of charge. It has no dependency on the path by which the charge is traveling. Therefore, if we move a charge from one point to ANOTHER in PRESENCE of an electric field in a straight line or a curve line, WORK done in both the cases will be the same i.e. potential difference between the points will be the same.

The CORRECT OPTION is (a) 28 × 10^-4 J

The explanation: The REQUIRED equation for the increase in ENERGY is given by:

U2 – U1 = \(\frac {1}{2}\) × C (V2^2 – V1^2)

= \(\frac {1}{2}\) × 8 × 10^-6 × (40^2 – 30^2)

= \(\frac {1}{2}\) × 8 × 10^-6 × 700

= 4 ×700 × 10^-6

= 2800 ×10^-6

= 28 × 10^-4 J

Therefore, the increase in energy is 28 × 10^-4 J.

The CORRECT option is (b) Decreases

Best EXPLANATION: When the dielectric slab is introduced between the plates, the induced surface charge on the dielectric reduces the electric field. The reduction in the electric field results in a DECREASE in POTENTIAL difference.

V = Ed = \(\frac {E_0 d}{K}=\frac {V_0}{k}\).

Correct ANSWER is (a) Permeability of the medium between the plates

The best explanation: The capacitance of a parallel PLATE CAPACITOR is directly proportional to the area of the plates and permittivity of the medium between the plates. It is INDIRECTLY proportional to the DISTANCE between the plates.

The CORRECT answer is (a) Equipotential

Explanation: ELECTRIC FIELD at any POINT is equal to the negative of the potential gradient. But inside a conductor, the electric field is zero. Hence, the electric potential is constant throughout the volume of a conductor and has the same VALUE on its surface. Thus the surface of a conductor is equipotential.

The CORRECT choice is (B) False

Easy explanation: In a thunderstorm accompanied by lightning, it is safest to SIT inside a car, RATHER than near a tree or on the open ground. The METALLIC body of the car becomes an electrostatic shielding from lightning.

The CORRECT CHOICE is (a) V=\(\frac {1}{4 \pi \varepsilon_o} (\frac {q_1}{r_1} + \frac {q_2}{r_2} + \frac {q_3}{r_3} +……. \frac {q_n}{r_n})\)

To explain I would say: The ELECTRIC POTENTIAL at a point due to a system of charges is EQUAL to the algebraic sum of the electric potentials due to individual charges at that point. The expression for electric potential due to a system of charges is given by:

V=\(\frac {1}{4 \pi \varepsilon_o} (\frac {q_1}{r_1} + \frac {q_2}{r_2} + \frac {q_3}{r_3} +……. \frac {q_n}{r_n})\)

The CORRECT ANSWER is (b) False

Explanation: In the inner region between the two capacitor plates, the electric fields DUE to the two CHARGED plates add up. The NET field is given by:

\(\frac {\sigma}{2\varepsilon_0}+\frac {\sigma}{2\varepsilon_0}=\frac {\sigma}{\varepsilon_0} \).

Correct answer is (d) Zero

The explanation is: The potential difference between any two POINTS on an EQUIPOTENTIAL surface is zero.

Work done=Test charge X potential difference(0)

=Zero .

The correct CHOICE is (d) Constant

To explain: Electric FIELD at any point is equal to the negative of the potential gradient. But INSIDE a CONDUCTOR, the electric field is ZERO. Hence, the electric potential is constant throughout the volume of a conductor and has the same value on its surface.

The correct choice is (d) 0V

The best I can EXPLAIN: Distance of the center of the triangle from EVERY corner point is = \(\FRAC {2}{3} * \frac {\SQRT 3}{2}\) * √3cm=1cm [as we know that the center divides the median in a 2:1 ratio and the length of the median is \(\frac {2}{3}\)*side length]. Therefore, the potential at the center is =\(\frac {5}{1} + \frac {5}{1} -\frac {10}{1}\)=0 esu Volt.