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Right choice is (c) 1 X 10^-2 J

The best explanation: Given: C1 = 6μF; V1 = 100 V

Initial energy stored (U1) = \(\FRAC {1}{2}\) C1V1^2 = \(\frac {1}{2}\) × 6 × 10^-6 × 100 × 100

= 3 × 10^-2 J

Potential (V) = \(\frac {C_1V_1}{(C_1+C_2)}\)

= 6 × 10^-6 × \(\frac {100}{(6+3)}\) × 10^-6

= \(\frac {600}{9}\) V

Final energy (U2) = \(\frac {1}{2}\) (C1 + C2) × V^2

= \(\frac {1}{2}\) × 9 × 10^-6 × \(\frac {1}{2} \times \frac {600}{9} \times \frac {600}{9}\)

= 2 × 10^-2 J

Therefore, the energy dissipated in the form of heat and electromagnetic radiation is:

U2 – U1 = 3 × 10^-2 – 2 × 10^-2

U2 – U1 = 1 × 10^-2 J

Right answer is (a) Increases

The EXPLANATION is: When a DIELECTRIC is introduced, and the battery REMAINS connected across the CAPACITOR, the capacitance increases from C0 to C.

C = kC0.

The CORRECT OPTION is (d) 8

Easy explanation: Capacitance with dielectric = 800 F

Capacitance without dielectric = 100 F

Dielectric constant = \( ( \FRAC {Capacitance \, with \, dielectric}{Capacitance \, without \, dielectric} ) \)

k = \(( \frac {800 F}{100 F} )\)

k = 8.

Therefore, the dielectric constant is calculated as 8.

Correct CHOICE is (c) 2

To ELABORATE: Capacitance without dielectric = 40 F.

Capacitance with dielectric = 80 F.

K = \(\frac {80}{40}\)

k = 2.

Correct ANSWER is (a) Proportional to each other

To elaborate: The electric field at the surface of a charged conductor is proportional to the surface charge density. The electric field is zero INSIDE the conductor and just outside, it is NORMAL to the surface. The contribution to the total flux comes only from its outer cross-section.

The correct option is (a) Zero

For explanation I would say: We know that electric potential does not depend on the path, i.e. it is a state function. THEREFORE if we rotate a charge AROUND another in a COMPLETE CIRCULAR path, the potential energy of its initial and final points is the same. Therefore, there is no change of potential energy of the charge and hence net work DONE on the charge is zero.

Correct choice is (b) False

For explanation: Van de Graff generators typically produce a very small amount of current, i.e. the current produced is in microamperes. THEREFORE, an accidental shock from a Van de Graff generator may be startling and it may be painful, but it will not cause SERIOUS HARM to most individuals, let alone KILL, even at a high voltages.

The CORRECT answer is (c) Remains constant

Best EXPLANATION: As the BATTERY remains connected ACROSS the capacitor, so the potential difference remains constant at V0 even after the introduction of the dielectric SLAB. In this way, dielectric has an effect on potential difference.

Right answer is (a) \(\frac {A\varepsilon_0}{d}\)

Easy 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.

C = \(\frac {Q}{V} = [ \frac {\sigma A}{(\frac {\sigma d}{\varepsilon_0})} ] = \frac {A\varepsilon_0}{d}\).

Correct choice is (B) Electrostatic shielding

Easy explanation: The PHENOMENON of making a region free from any electric field is CALLED electrostatic shielding. It is based on the FACT that the electric field vanishes inside the CAVITY of a hollow conductor.

Right choice is (d) ML^2T^-3A^-1

The EXPLANATION: Electrostatic potential energy is a scalar QUANTITY that means it possesses only MAGNITUDE and no DIRECTION. This quantity is denoted by U and it is measured in joules (J). The dimensional formula of electrostatic potential energy is ML^2T^-3A^-1.

Correct option is (a) 36V

The explanation is: Distance of the center of the SQUARE from the corner points=√2*\(\frac {\SQRT 2}{2}\)m=1m. Therefore, electric potential at the center of the square due to the charges = V=\(\frac {1}{4\pi \epsilon} \frac {(2*10-9+1*10-9-2*10-9+3*10-9)}{1}\)=9*10^9*(2*10-9+1*10-9-2*10-9+3*10-9)=9*(2+1-2+3) =36V.

Correct answer is (a) \(\frac {q}{r}\)

EASY explanation: FORCE on a unit point charge kept at a distance r from the charge=\(\frac {q}{r^2}\). Therefore, WORK done to BRING that point charge through a small distance DR=\(\frac {q}{r^2}\)*(-dr). Therefore, the potential of that point is =\(\int_\alpha^r\frac {-q}{r^2}dr = \frac{q}{r}\).

The correct ANSWER is (d) VA=VB

Easy explanation: We KNOW thatVAB=VA-VB=-\(\int_A^B\)E.dx. But E=0, so the value of the integral becomes zero. THEREFORE VA=VB is obtained. The electric field is ALWAYS directed from a point of higher potential to a point of LOWER potential. But if the potential of two points is the same, i.e. no potential difference, then there will be no electric field.

The correct choice is (a) 1.6*10^-19

For explanation I would SAY: 1 electron volt is the amount of work done if an electron is passed through a potential difference of 1V. Therefore the work done = 1V*charge of an electron = 1.602*10^-19 J. But it is a small quantity and hence we USE kilo electron volt and MEGA electron volt in PRACTICAL.

Correct choice is (c) Remains unchanged

Best explanation: As the POTENTIAL difference remains unchanged, so the electric FIELD E0 between the CAPACITOR plates remain unchanged.

E = \(\frac {V}{d}\) = \(\frac {V_0}{d}\) = E0.

The CORRECT CHOICE is (c) Capacitance increases

For explanation I would say: When a dielectric material is inserted between the plates of the parallel PLATE capacitor, the capacitance of the capacitor increases with a factor of K. That is C=KC0.

The correct answer is (a) 1.8F

To EXPLAIN I would SAY: SINCE we know that capacitance=\(\FRAC {charge}{voltage}\).

Therefore capacitance=\(\frac {9}{5}\)=1.8F.

Right ANSWER is (d) Polarisation density

To elaborate: The induced DIPOLE moment developed PER unit volume of a dielectric when placed in an external electric FIELD is called polarization density. It may be defined as the charge induced per unit surface area.

Right answer is (a) Increases

For explanation: When the battery is disconnected, and a DIELECTRIC is INTRODUCED, there will be a decrease in potential difference and as a result, the CAPACITANCE increases k times.

C = \(\FRAC {Q_0}{V}\)

C = \( [ \frac {Q_0}{(\frac {V_0}{k})} ] \)

C = \(\frac {kQ_0}{V_0}\)

C = kC0.

The correct answer is (b) 1.125*10^12V

The explanation: In the SI system, electric potential due to a point charge at a distance r is \(\frac {Q}{4\pi \VAREPSILON r}\)=9*10^9*\(\frac {q}{r}\). Substituting the values, we GET potential=9*10^9*\(\frac {10}{0.08}\)V=1.125*10^12V. Though in practice, this HUGE value of electric potential is not present.

The correct option is (b) Both will have the same energy

The BEST I can EXPLAIN: Electrostatic potential energy of Q1 → U1 = \(\frac {1}{(4\pi \varepsilon_o)}\) × [Q × \(\frac {q}{r}\)] ……….1

Electrostatic potential energy of Q2 → U2 = \(\frac {1}{(4\pi \varepsilon_o)}\) × [Q × \(\frac {2q}{2r}\)]

→ U2 = \(\frac {1}{(4\pi \varepsilon_o)}\) × [Q x \(\frac {q}{r}\)] ………2

1 = 2

Therefore, both the charges will have the same energy.

Correct CHOICE is (b) False

The explanation is: ELECTRIC FIELD lines cut the equipotential surfaces at an angle of 90 degrees. The direction of the electric field is the same as the direction of electric field lines. Therefore, the field is perpendicular to the equipotential POINTS, not tangent to them.

Correct CHOICE is (a) Zero

Easy explanation: We CONSIDER earth as the storage of infinite POSITIVE as WELL as a negative CHARGE. Therefore, the potential of the earth is always considered to be zero and the potential of every body is measured with respect to earth. That’s why if we connect any charged body to the earth, its potential instantaneously becomes zero.

Correct ANSWER is (C) Depends on the MAXIMUM electric field

Easiest explanation: The maximum energy that can be desirably stored in a CAPACITOR is depends on the maximum electric field that the DIELECTRIC can withstand without breaking down. Therefore, capacitors of the same type have about the same maximum energy density, i.e. joules of energy per cubic meter.

Right option is (c) U = \(\frac {1}{2}\) (QV)

To EXPLAIN: Work DONE in charging a capacitor gets stored in the capacitor in the form of its electric potential ENERGY and this is known as energy stored in a capacitor. The expression is given by:

U = \(\frac {1}{2}\) (CV^2) = \(\frac {1}{2}\) (QV) = \(\frac {1}{2} (\frac {Q^2}{C})\)

The unit is for this QUANTITY is joule.

Correct answer is (b) PARALLEL PLATE capacitor

The best I can explain: The simplest and the most widely used capacitor is the parallel plate capacitor. It consists of two large plane parallel conducting plates, separated by a SMALL distance.

Capacitance of a parallel plate capacitor = \(\frac {Q}{V} = [ \frac {\sigma A}{(\frac {\sigma d}{\varepsilon_0})} ] = \frac {A\varepsilon_0}{d}\).

The correct answer is (b) Uniform

Easiest explanation: The DIRECTION of the electric field is from the positive to the negative PLATE. In the inner region, between the TWO capacitor PLATES, the electric fields due to the two charged plates add up. HENCE, the field is uniform throughout.

The CORRECT choice is (a) True

Easiest explanation: Yes, both are related to electric potential. The RELATIONSHIP between electric potential and electric field is a differential → electric field (E) is the gradient of electric potential (V) in the x direction. Thus, as the CHARGE moves in the x direction, the rate of its change in potential is the value of the electric field. Then, electric field can be defined as the NEGATIVE of the rate of DERIVATIVE of potential difference.

The correct option is (a) 1N/C

For explanation I would SAY: From the definition, we KNOW that electric field E=\(\FRAC {-dv}{dx}\). Therefore V/m is the unit of electric field intensity. 1 V/m MEANS the amount of electric field in which if we move a unit positive charge by 1 m, the work done will be 1N. Therefore, the electric field will be 1N/C.

The correct option is (b) False

The explanation: When a dielectric is INTRODUCED, and he BATTERY remains connected across the CAPACITOR, the charge on the capacitor PLATES increases from Q0 to Q.

Q = CV = kC0.V0 = kQ0.

The correct ANSWER is (C) Zero

To explain: In the outer regions above the upper plate and below the lower plate, the electric fields due to the two CHARGED PLATES cancel out. Hence, the NET electric field in the outer regions above the plate and below the lower plate is zero.

The correct option is (c) Non-polar

Easiest explanation: A MOLECULE in which the centre of mass of positive and negative charges collide with each other is CALLED a non-polar molecule. They normally have ZERO DIPOLE moment. They have symmetrical shapes.

Correct option is (b) 2q

The BEST explanation: Electric potential due to the CHARGES at the origin= q*\((\frac {1}{1} + \frac {1}{2} + \frac {1}{4} + \frac {1}{8} +…..=q*\frac {1}{1-\frac {1}{2}})\)=2q [by applying the formula of infinite G.P. series]. It can be shown that if the charges are alternately positive and negative, then the potential at the CENTER would be \(\frac {2q}{3}\) .

The correct choice is (c) Zero

Explanation: By Gauss’s THEOREM, the charge enclosed by the gaussian surface is zero. Consequently, the ELECTRIC field MUST be zero at EVERY point INSIDE the cavity. Then, the entire excess charge lies on its surface.

The correct option is (d) x=+\(\sqrt{\frac {a}{3b}}\) and x=-\(\sqrt{\frac {a}{3b}}\)

To EXPLAIN: We KNOW that E=-\(\frac {dv}{dx}\)

∴ E = –\(\frac {d}{dx}\) (ax – bx^3) = -(a – 3bx^2)

Electric FIELD will be ZERO. ∴ E = 0 ⇒ (a-3bx^2) = 0

⇒ a = 3bx^2

⇒ x^2 = \(\frac {a}{3b}\) = x = ±\(\sqrt{\frac {a}{3b}}\)

Correct choice is (b) 0.5 m

For explanation I WOULD SAY: Electric field DUE to a charge q at a distance r is \(\frac {q}{kr^2}\)

 and electric potential at that point is \(\frac {q}{kr}\). Therefore, \(\frac {q}{kr^2}\)=32 and \(\frac {q}{kr}\)=16. Dividing these TWO EQUATIONS, we get r=0.5m. If the medium is air, k=1. Thus we can get the value of q=0.89 C.

Correct choice is (c) 4.425 × 10^-12 F

To explain I would say: C = \(\frac {Q}{V} = [ \frac {\sigma A}{(\frac {\sigma d}{\varepsilon_0})} ] = \frac {A\varepsilon_0}{d}\).

C = \(\frac {100 \times 10^{-4} \times 8.85 \times 10^{-12})}{(20 \times 10^{-3})}\)

C = 4.425 × 10^-12 F.

The correct option is (b) M^-1L^-2T^4I^2

For explanation I would say: We know that CAPACITANCE=CHARGE*(voltage)^-1

Now charge=current*time=[I^1T^1].

voltage=electric FIELD*distance, and electric field=FORCE*(charge)^-1

Since we know the dimensional formula of force is [M^1L^1T^-2], therefore,

Electric field=[M^1L^1T^-2] * [I^1T^1]^-1=[M^1L^1T^-3I^-1].

Finally, voltage becomes=[M^1L^1T^-3I^-1] *L.

Hence, we conclude that dimensional formula of capacitance is [I^1T^1] * ([M^1L^1T^-3I^-1] *L)^-1 that is M^-1L^-2T^4I^2.

Correct choice is (a) True

For explanation I would say: If the CHARGE ‘Q’ is negative, the sign should be considered in the equation. Therefore, a SYSTEM consisting of negative and positive point charges will have negative potential energy. A negative potential energy means that work MUST be done against the electric field in order to move the charges apart.

Correct CHOICE is (b) 0 V

Explanation: Electric potential is a scalar quantity. So, the total potential will be the SUM of all the individual charges.

V1 = \(\FRAC {kQ}{r} = \frac {2k}{3}\)

V2 = \(\frac {kQ}{r} = \frac {-2k}{3}\)

Total potential (V) = V1 + V2

V = K [ \((\frac {2}{3}) + (\frac {-2}{3})\) ]

V = 0 V

Therefore, the potential at the midpoint between the two charges is zero volt.

Right CHOICE is (b) False

The EXPLANATION: Electric potential is a positive QUANTITY if both the charges are positive. In this case, all the three charges are positive; hence there is no negative term in the total energy term. Therefore, we cannot make the total energy of the system zero by adjusting the VALUES of charge. But it would be possible if one of the charges were positive.

Correct answer is (a) 0.8J

The best I can EXPLAIN: Work done = potential*charge by DEFINITION. We know that the potential of a POINT is the amount of work done to bring a UNIT charge from infinity to a certain point. Therefore, work done W=q*V=4*10^-3*200J=0.8J. The work done is positive in this case.

The correct answer is (c) Zero

For explanation I would say: ELECTRIC POTENTIAL is a scalar quantity and its value is solely dependent on the charge near it and the DISTANCE from that charge. In this case, the point is EQUIDISTANT from the two point charges and the point charges have the same value but opposite nature. Therefore equal but opposite potentials are generated due to the charges and hence the net potential at midpoint BECOMES zero.

Right answer is (c) [M L T^-3 I^-1]

To explain: Potential can be simply defined as work done on a unit charge, therefore the DIMENSION=\(\FRAC {the \, dimension \, of \, work}{the \, dimension \, of \, charge}=\frac {[M L T^{-2}]}{[I T]}\)=[M L T^-3 I^-1]. The dimension of potential energy in MECHANICS is not the same as ELECTRIC potential energy though both of them are the unit of energy. In mechanics, it is defined as work done on a unit mass.