1638 + Interview Questions in ELECTROSTATICS IN GENERAL AWARENESS Page 3 InterviewSolution

101.	An uncharged parallel-plate capacitor filled with a material of dielectric constant k is connected to a parallel-plate air capacitor of identical geometry charged to a potential V. At equilibrium, common potential difference across them is V’. The dielectric constant k is equal to V’-V(A) \(\cfrac{V'-V}{V'}\)(B) \(\cfrac{V'-V}{V'+V}\)(C) \(\cfrac{V-V'}{V'}\)(D) \(\cfrac{V+V'}{V-V'}\)
Answer» Correct option is (C) \(\cfrac{V-V'}{V'}\) [Hint : Common potential, V' = \(\cfrac{total\,charge}{total\,capacitance}\) = \(\cfrac{CV}{KC+C}\)]

101.

An uncharged parallel-plate capacitor filled with a material of dielectric constant k is connected to a parallel-plate air capacitor of identical geometry charged to a potential V. At equilibrium, common potential difference across them is V’. The dielectric constant k is equal to V’-V(A) \(\cfrac{V'-V}{V'}\)(B) \(\cfrac{V'-V}{V'+V}\)(C) \(\cfrac{V-V'}{V'}\)(D) \(\cfrac{V+V'}{V-V'}\)

Answer»

Correct option is (C) \(\cfrac{V-V'}{V'}\)

[Hint : Common potential, V' = \(\cfrac{total\,charge}{total\,capacitance}\) = \(\cfrac{CV}{KC+C}\)]

102.	The energy stored in a charged capacitor is U. The capacitor is isolated and connected across the terminals of an identical uncharged capacitor. The energy stored in each capacitor is (A) U (B) 3U/4 (C) U/2 (D) U/4
Answer» Correct option is (D) U/4

Discussion

103.	Three capacitors C1, C2 and C3 are connected to a battery of p.d. V as shown. Which of the following are the correct relations for the charges on the capacitors and the p.d.s across them ?(A) Q1 = Q2 = Q3 and V1 = V2 = V3 = V (B) Q1 = Q2 + Q3 and V = V1 + V2 + V3 (C) Q1 = Q2 + Q3 and V= V1 + V2(D) Q2 – Q3 and V2 = V3
Answer» (C) Q₁ = Q₂ + Q₃ and V= V₁ + V₂

Discussion

104.	A copper plate of thickness b is inserted between the plates of a parallel-plate capacitor of plate separation d. If b = d/3, the capacitances before and after the insertion of the plate are in the ratio(A) 2 : 3 (B) 3 : 2 (C) 1 : √3(D) √3 : 1.
Answer» Correct option is (A) 2 : 3

Discussion

105.	Parallel plate capacitor is constructed using three different dielectric materials as shown in the figure. The parallel plates, across with a potential difference is applied of area A `"metre"^(2)` and separated by a distance d metre. The capacitance across A and B is A. `(epsilon_(0)A)/(d)[(K_(1))/(2)+(K_(2)K_(3))/(K_(2)+K_(3))]`B. `(epsilon_(0)A)/(d)[(K_(1))/(2)+(K_(2)+K_(3))/(K_(2)K_(3))]`C. `(epsilon_(0)A)/(d)[(2)/(K_(1))+(K_(2)K_(3))/(K_(2)+K_(3))]`D. `(epsilon_(0)A)/(d)[(2)/(K_(1))+(K_(2)+K_(3))/(K_(1)K_(3))]`
Answer» Correct Answer - A The resultant will be `(epsilon_(0)A)/(d)[(K_(1))/(2)+(K_(2)K_(3))/(K_(2)+K_(3))]`

Discussion

106.	A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers `1//3` of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is `C_1`. When the capacitor is charged, the plate area covered by the dielectric gets charge `Q_1` and the rest of the area gets charge `Q_2`. The electric field in the dielectric is `E_1` and that in the other portion is `E_2`. Choose the correct option/options, ignoring edge effects. A. `(E_(1))/(E_(2)) = 1`B. `(E_(1))/(E_(2)) = (1)/(K)`C. `(Q_(1))/(Q_(2)) = (3)/(K)`D. `(C_(1))/(C_(2)) = (3 + K)/(K)`
Answer» Correct Answer - A When capacitor is charged, both parts of capacitors have common potential difference `V`. So `E_(1) = (V)/(d) = E_(2)` or `(E_(1))/(E_(2)) = 1` `C_(1) = *K (in_(0) A//3)/(d) , C_(2) = (in_(0) 2 A//3)/(d)` `C = C_(1) + C_(2) = (K in_(0) A)/(3d) + (2 in_(0) A)/(3d) + ((k-2) in_(0) A)/(3d)` `(C )/(C_(1)) = (k + 2)/(K)` `Q_(1) = C_(1) V = (K in_(0) A)/(3d)` `Q_(2) = C_(2)V = (2 in_(0) A)/(3d) V` `(Q_(1))/(Q_(2)) = (K)/(2)`

Discussion

107.	A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process isA. (a) zeroB. (b) `1/2(K-1)CV^2`C. (c) `(CV^2(K-1))/(K)`D. (d) `(K-1)CV^2`
Answer» Correct Answer - A The potential energy of a charged capacitor before removing the dielectric slat is `U=(Q^2)/(2C)`. The potential energy of the capacitor when the dielectric slat is first removed and the reinserted in the gap between the plates is `U=(Q^2)/(2C)` There is no change in potential energy, therefore work done is zero.

Discussion

108.	A metal foll of negative thickness is intorduced between two plates of a capacitor at the center. What will be the new capacitance of the capacitance ?
Answer» On intorducing a thin metal foil d is hallved, Arrangment is equivalent to two conderers each of capacity 2C in series. `:. (1)/(C_(s)) = (1)/(2C) + (1)/(2C) = (1)/(C ) :. C_(s) = C` Capacity is uncharged.

Discussion

109.	If the potential of a capacitor having capacity of `6 muF` is increased from 10 V to 20 V,then increase in its energy will beA. `4 xx 10^(-4)J`B. `4 xx 10^(-14)J`C. `9 xx 10^(-4)J`D. `12 xx 10^(-6)J`
Answer» Correct Answer - C

Discussion

110.	On charging a parallel - plate capacitor to a potentia V, the spacing between the plates is halved and a dielectric medium of `in_(r) = 10` is introcded between the paltes, without disconnecting the dc source. Explain using suitable expression how the (a) capacitance (b) electric field (c ) energy density of the capacitor change.
Answer» As the d.c, source remains connected p.d., (V) between the plates of capacitor remains uncharged even after dielectric is insered between the plates. (a) Origanal capacitance `C_(0) = (in_(0) A)/(d)` New capacitance `C_(0) = (varepsilon_(r) in_(0) A)/(d//2) = 20 C_(0)` (b) Changed electric field, `E = (V)/(d//2) = 2 (V//d) = 2 E_(0)` (c ) Origanal energy density, `U = (1)/(2)" in "E^(2)` `= (1)/(2) (varepsilon_(r) varepsilon_(0)) (2E_(0))^(2)` `xx varepsilon_(0) E_(0)^(2))` `= 4xx10 U_(0) = 40 U_(0)`

Discussion

111.	A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process isA. zeroB. `(1)/(2) (K - 1) CV^(2)`C. `(CV^(2) (K - 1))/(K)`D. `(K - 1) CV^(2)`
Answer» Correct Answer - A Initial energy `U_(i) = (Q^(2))/(2 K C)` When the slab is removed, `U = (Q^(2))/(2C)`, When the slab is removed, `U = (Q^(2))/(2C)`, When dielectric slab is introduced, energy becomes `U_(f) = (Q^(2))/(2 KC) = U_(i)`. `:.` Net work done by the system `= U_(f) - U_(i)` = Zero

Discussion

112.	A parallel plate capacitor is maintained at a certain potentail difference. When a 3mm thick slab is intorduced between the plate, in a order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.
Answer» Here, `t = 3mm, x = 2.4 mm, K = ?` With air as dielectric, `C_(0) = (in_(0) A)/(d)` If `d_(1)` is new distance of separation between the plates when dielectric is introduced, then `C_(0) = (in_(0) A)/(d_(1) - t (1- (1)/(K)))` As `C = C_(0)` , therefore , `d_(1) - t (1- (1)/(K)) = d` or `t (1 - (1)/(K)) = d_(1) - d = x` As t = 3mm and `x = d_(1) -d = 2.4 mm` `:. 3(1 - (1)/(K)) = 2.4` `1 - (1)/(K) = (2.4)/(3) = 0.8` `(1)/(K) = 1-0.8 = 0.2, K = `

Discussion

113.	In the circuit shown in Fig. each capacitor has a capacity of `3 muF`. Calculate the quantity fo charge on each capacitor.
Answer» Total resitance is the circuit ABCD, `R = 4+1 = 5 Omega`. `:.` Current `I = (V)/(R ) = (10)/(5) = 2A` Potential diff. across A and B `= I xx 4 = 2xx4 = 8V` As two capacitors of `3 muF` each are in series, `:.` pot. Diff across each condenser `= (8)/(2) = 4V` charge on each condenser, `q = CV = 3xx4` `= 12 muC`

Discussion

114.	A parallel plate capacitor of 6 `mu F ` is connected across 18 V battery and charged. The battery is k = 2.1 is introduced between the plates. What will be the charge on capacitor ?A. `50muC`B. `108muC`C. `60muC`D. `85muC`
Answer» Correct Answer - B `C=6muF,V=18V,k=2.1,Q=?` `Q=C_(m)*V=C_("air")kV` `=6xx10^(-6)xx2.1xx18=108xx10^(-6)C`

Discussion

115.	A parallel plate condenser with air as dielectric has capacity `C_(0)=(in_(0)A)/d`. A thin mica sheet of dielectric constant K and thickness t is introduced near the first plate and then moved with constant velocity v towards the other plate. The capacity of the condenser will beA. `(in_(0)A)/(d-t+t/k)`B. `(in_(0)Av)/(d-t+t/k)`C. `(in_(0)Av)/(d-t)`D. `(in_(0)A)/(d-t)`
Answer» Correct Answer - A The capacity of parallel plate condenser does not depend upon the velocity of the plate. Thus, if mica sheet of thickness t and dielectric K is introduced between the plates of parallel plate condenser, then its capacity is given by, `(in_(0)A)/(d-t+t/k)`

Discussion

116.	A parallel plate capacitor contanins a mica sheet (thickness ` 0. 5+xx10^(-3) m`) . And a sheet of fiber (thickness `0. 5 xx 10^(-3) m`) . The dielectric constant of mica is `8` and that of thye fiber is `2.5` Assuming that the fiber breaks down when subjected to an electric field of ` 6.4 xx 10^6 Vm^(-1)`. , find the maximum safe voltage that can be applied to the capacitor.
Answer» Let `sigma` be the surface charge density of capacitor plates. for mica `E_(1) = (sigma)/(K_(1) in_(0))` and for fibre `E_(2) = (sigma)/(K_(2) in_(0)) or (E_(1))/(E_(2)) = (K_(2))/(K_(1))` As `E_(2) = 6.4xx10^(6) V//m` `E_(1) = (K_(2))/(K_(1)) xx E_(2) = (2.5)/(8) xx 6.4xx10^(6) = 2xx10^(6) V//m` Maximum voltage for capacitor `V = E_(1) d_(1) + E_(2) d_(2) = 2xx10^(3) + 3.2xx10^(3) = 5200V`

Discussion

117.	The capacity of a parallel plate condenser is C. When the distance between the plates is halved, its capacity isA. 2 CB. CC. 0.25 CD. 0.2 C
Answer» Correct Answer - A `d_(2)=d_(1)/2,C_(2)=?,C_(1)=C` `Cprop1/d` `Now" "C_(2)/C_(1)=d_(1)/d_(2)` `C_(2)=(C_(1)d_(1))/d_(2)=(Cd_(1))/(d_(1)/2)=2C`

Discussion

118.	A parallel plate condenser consists of two plates each of area `100cm^(2)`. They are separated by mica sheet of thickness 8.85 mm. If the relative permittivity of mica is 6. Then capacity of parallel plate condenser isA. 50 pFB. 40 pFC. 60 pFD. 30 pF
Answer» Correct Answer - C `C=(Akepsi_(0))/d=(100xx10^(-4)xx6xx8.85xx10^(-12))/(8.85xx10^(-3))` = 60 pF.

Discussion

119.	The capacity of a parallel plate condenser increases four times if the air between the plates is replaced by glass. The permittivity of glass will beA. `3.54xx10^(-11)C^(2)//Nm^(2)`B. `3.54xx10^(-12)C^(2)//Nm^(2)`C. `2.54xx10^(-11)C^(2)//Nm^(2)`D. `2.22xx10^(-12)C^(2)//Nm^(2)`
Answer» Correct Answer - A `C_(m)=C_("air")xxk" "thereforek=4` `in=in_(0)k=8.85xx10^(-12)xx4` `=3.54xx10^(-11)C^(2)//Nm^(2)`

Discussion

120.	The capacity of a parallel plate condenser is `12muF`. Its capacity, when the separation between plates is doubled and area is halved will beA. `3muF`B. `12muF`C. `6muF`D. `1.5muF`
Answer» Correct Answer - A `C_(1)=12muF, C_(2)=?` If `d_(2)=2d_(1)andA_(2)=A_(1)/2` `CpropA/d` `therefore" "C_(2)/C_(1)=A_(2)/A_(1)xxd_(1)/d_(2)=A_(1)/(2A_(1))xxd_(1)/(2d_(1))=1/4` `C_(2)=C_(1)/4=12/4=3muF`

Discussion

121.	An electric dipole in a uniform electric field experiences (When it is placed at an angle `theta` with the field)A. no net force and no torqueB. a net force but a torqueC. a net force and a torqueD. no net force but a torque
Answer» Correct Answer - B::C::D

Discussion

122.	To reduce the capacity of a parallel plate condenser, separation between the plates isA. reduced and area of the plates decreasedB. decreased and area of the plates increasedC. increased and area of the plates decreasedD. increased and area of the plates increased
Answer» Correct Answer - C

Discussion

123.	An electric dipole in a uniform electric field experiences (When it is placed at an angle `theta` with the field)A. force onlyB. torque onlyC. both force and torqueD. neither a force nor a torque
Answer» Correct Answer - B

Discussion

124.	Four point charges `-Q, -q, 2q` and `2Q` are placed, one at each corner of the square. The relation between `Q` and `q` for which the potential at the centre of the square is zero isA. `Q =- q`B. `Q =- (1)/(q)`C. `Q = q`D. `Q = (1)/(q)`
Answer» Correct Answer - A

Discussion

125.	An electric dipole is kept in non-unifrom electric field. It experiencesA. a force and a torqueB. a force but not a torqueC. a torque but not a forceD. Neither a force nor a torque
Answer» Correct Answer - A

Discussion

126.	A parallel plate condenser has a unifrom electric field `E (V//m)` in the space between the plates. If the distance between the plates is `d(m)` and area of each plate is `A(m^(2))` the energy (joule) stored in the condenser isA. `(1)/(2)epsilon_(0)E^(2)`B. `epsilon_(0)EAd`C. `(1)/(2)epsilon_(0)E^(2)`D. `E^(2)Ad//epsilon_(0)`
Answer» Correct Answer - C

Discussion

127.	In a region, the potential is respresented by `V(x, y, z) = 6x - 8xy - 8y + 6yz`, where `V` is in volts and `x, y, z` are in meters. The electric force experienced by a charge of `2` coulomb situated at point `(1, 1, 1)` isA. `6 sqrt(5) N`B. `30 N`C. `24 N`D. `4 sqrt(35) N`
Answer» Correct Answer - D Given, `V = 6x - 8 xy - 8 y + 6 yz` `(E_(x))_(1,1,1) = -(del V)/(dx) = -(6-8 y) = -(6-8xx1) = 2` `(E_(y))_(1,1,1) = -(del V)/(dy) = - (-8x -8 + 6z)` `= -(-8xx1-8+6xx1) = 10` `(E_(z))_(1,1,1) = -(del V)/(dz) = -(6y) = -6xx1=-6` `E = sqrt(E_(x)^(2) +E_(y)^(2) + E_(z)^(2)) = sqrt(2^(2) +10^(2) +(-6)^(2))` `= sqrt(140) = 2 sqrt(35) N//C` `F = qE = 2xx2 sqrt(35) = 4 sqrt(35) N`

Discussion

128.	`A, B` and `C` are three points in a unifrom electric field. The electric potential is A. maximum at AB. maximum at BC. maximum at CD. same at all the three points A,B and C
Answer» Correct Answer - B

Discussion

129.	A conducting sphere of radius `R` is given a charge `Q`. The electric potential and the electric field at the centre of the sphere respectively areA. zero and `(Q)/(4pi epsilon_(0)R^(2))`B. `(Q)/(4piepsilon_(0)R)` and zeroC. `(Q)/(4pi epsilon_(0)R)` and `(Q)/(4pi epsilon_(0)R^(2))`D. Both are zero
Answer» Correct Answer - B

Discussion

130.	The electric potential V at any point x,y,z (all in metre) in space is given by `V=4x^2` volt. The electric field at the point `(1m, 0, 2m)` is ……………`V/m`.A. 8, along negative X-axisB. 8, along positive X-axisC. 16, along negative X-axisD. 16, along positive Z-axis
Answer» Correct Answer - A Electric field, `E=-(dV)/(dr)=-(d)/(dx)(4x^(2))=-8x` `=-8(1)=-"8 Vm"^(-1)` Negative sign indicates E is along negative direction of X-axis.

Discussion

131.	In a region, the potential is respresented by `V(x, y, z) = 6x - 8xy - 8y + 6yz`, where `V` is in volts and `x, y, z` are in meters. The electric force experienced by a charge of `2` coulomb situated at point `(1, 1, 1)` isA. `6sqrt(5)N`B. `30N`C. `24N`D. `4sqrt(35)N`
Answer» Correct Answer - D

Discussion

132.	The work done in placing a charge of `8xx10^-18` coulomb on a condenser of capacity 100 micro-farad isA. (a) `16xx10^-32` jouleB. (b) `3.1xx10^-26` jouleC. (c) `4xx10^-10` jouleD. (d) `32xx10^-32` joule
Answer» Correct Answer - D The work done is stored as the potential energy. The potential energy stored in a capacitor is given by `U=1/2(Q^2)/(C)=1/2xx((8xx10^-18)^2)/(100xx10^-6)=32xx10^-32J`

Discussion

133.	For the situation shown in the figure below ("assume "`r gtgt` length fo dispole ) mark out the correct statement (s) - .A. Force acting on the dipole is zeroB. Force acting on the dipole is approximately ` (PQ)/(4 pie_(0) r^(2))` and is acting upwardC. Torque acting on the dipole is ` (pq)/(4 pi e_(0)r^(2))` clockwise directionD. Torque acting on the dipole is `(pQ)/(4 pi e_(0)r^(2))` in ant i-clockwise direction
Answer» Correct Answer - B::C (A) Force on charge due to diple `vec F =Qvec E rArr = Q. 1/(4 pi in0) p/r^3` ` F= (pQ)/(4 pi in_0 r^3)` acting downward so force on diple will be equal and oppositede (C ) `vec tau = vec P xx P ( hat j ) xx 1/(4 pi in_0) Q/r^2 hat i` `vec tau = 1/(4 pi in0) (pQ)/r^3 (-hat k)`.

Discussion

134.	A conducting sphere of radius r has a charge . Then .A. the charge is uniformly distributed over its surface if there is an external electric field .B. Distribution of charge over its surface will be non uniform if no external electric field exist in space,C. Electric field strength inside the sphere will be equal to zero only when no external electric field existsD. Potential at every point of the sphere must be same
Answer» Correct Answer - D Evergy point on the sphere will have same potential .

Discussion

135.	A particel of mass m and charfe `q` is throun in a region wihere unirfrom gravittional field and electric field are present . The path of particle -A. may be a straight lineB. may be a circleC. may be parbolaD. may be a hyperbola
Answer» Correct Answer - A::C Path will be straight when gravitational and electric field are perpendicular.

Discussion

136.	A proton and a ionised deuterium are initially at rest and are acelerated through the same potential disfference. Which of the following is false concerning the final propertice of the two particels ?A. They have different speedsB. They have same momentumC. They have same kinetic energyD. They have been subjected to same force
Answer» Correct Answer - A::C `Delta K = K qV` `K= 1/2 mV^2` bothe have same charge so their ` KE` is same while their mass is different so their speed is different.

Discussion

137.	A particel of charge ` 1 mu C & ` mass `1 mg` moving with a velocity of ` 4 m//s` is subjected to a unifrm electric field of magnitude ` 300 V m` for ` 10 sece`. Then it s final speed cannot be :A. ` 0. 5 m//s`B. ` 5 m//s`C. ` 3 m//s`D. ` 6 m//s`
Answer» Correct Answer - A::B::C `a= (qE)/m = 0. 3 m//s^2 rArr at = 3 m//s^2` ` vec V = vec x + vec at` From vector addition `(x-at ) le V le u + a t`.

Discussion

138.	Figure shows some of the electric field lines due to three point charges `q_, q_2` and `q_3` of equal magnitude. What are the signs of each of the three charges?
Answer» Correct Answer - A::B::C::D Electric field lines start from positive charge and terminate of negative charge

Discussion

139.	A charge `q = -2.0 muC` is placed at origin. Find the electric field at `(3 m, 4 m, 0)`.
Answer» Correct Answer - A::B::C::D `E=(1/(4piepsilon_0)) (q/r^3)(r_p-r_q)` Here, `r=sqrt((3)^2+(4)^2)=5m` `:. E=((9xx10^9)(-2xx10^-6))/((5)^3)(3hati+4hatj)` `=-(4.32hati+5.76ecj)xx10^2N//C`

Discussion

140.	Variation of electrostatic potential along the x - direction is shown in (Fig. 3.142). The correct statement about electric field is .A. `x` comppnent at point B is maixmumB. `x` component at point A is towards positive x-axis .C. `x` component at point C is along positive x-axisD. `x` component at point C is along positive x-axis
Answer» Correct Answer - D ` E_x =- (dV)/(dx)=` Negative of sloope of V-x graph.

Discussion

141.	A uniform electric field of `400 V/m` is directed at `45^@` above the x-axis as shown in the figure. The potential difference `V_A-V_B` is given by
Answer» Correct Answer - D `E=400 cos45^@hati+400sin45^@hatj` `V_A-V_B=-int_B^AE.dr` where `dr=dxhati+dyhatj`

Discussion

142.	The electric field in a certain region is given by `E=(5hati-3hatj)kV//m`. Find the difference in potential `V_B-V_A`. If A is at the origin and point B is at a. (0,0,5)m, b. (4,0,3) m.`
Answer» Correct Answer - B Apply `V_B-V_A=-int_A^BE.dr` E is given in the question and `dr=dxhati+dyhatj` `:. E.dr=(5dx-3dy)` `:. -intE.dr=(3y-5x)` With limits answer answer comes out to be `V_B-V_A=3(y_f-y_i)-5(x_f-x_i)`

Discussion

143.	In figure two positive charges `q_(2)` and `q_(3)` fixed along the y-axis ,exert a net electric force in the `+x` direction on a charge `q_(1)` fixed along the x-axis if a positive charge `Q` is added at `(x,0)` the force on `q_(1)` A. shall increase along the positive X-axisB. shall decrease along the positive X-axis.C. shall point along the negative X-axisD. shall increase but the direction charges because of the intersection of Q with `q_(2)` and `q_(3)`
Answer» Correct Answer - A

Discussion

144.	Two points A and B are 2 cm apart and a uniform electric field E acts along the straight line `AB` directed A to B with `E=200N//C`. A particle of charge `+10^-6C` is taken from A to B along AB, Calculate a. the force on the charge b. the potential difference `V_A -V_B` and c.the work done on the charge by E
Answer» Correct Answer - A::B::D a. Electrostatic force on the charge, `F=qE=(10^-6)(200)` `=2xx10^-4 N` b. In uniform electric field, `PD. V=E.d` or `V_A-V_B=200xx2xx10^-2` `=4V` c. `W(2xx10^-4)(2xx10^-2)cos0^@` `=4xx10^-6J`

Discussion

145.	Four charges `Q_(1), Q_(2), Q_(3)` and `Q_(4)` of same magnitude are fixed along the x axis at `x=-2a, -a, +a` and `+2a`, respectively. A positive charge q is placed on the positive y axis at a distance `b gt 0`. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below the lists. `{:(,"List-I",,,"List-II"),((P),Q_(1)","Q_(2)","Q_(3)","Q_(4) "all positive",,(1),+x),((Q),Q_(1)","Q_(2) " positive, "Q_(3)","Q_(4)" negative",,(2),-x),((R),Q_(1)","Q_(4)" positive ,"Q_(2)","Q_(3)" negative",,(3),+y),((S),Q_(1)","Q_(3)" positive , "Q_(2)","Q_(4)" negative",,(4),-y):}`A. P-3, Q-1, R-4, S-2B. P-4, Q-2, R-3, S-1C. P-3, Q-1, R-2, S-4D. P-4, Q-2, R-1, S-3
Answer» Correct Answer - A

Discussion

146.	Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides d. Two of the point charges are identical and have charge q.If zero net work is required to place the three charges at the corners of the triangles, what must the value of the third charge be?
Answer» Correct Answer - B U=0 `:. k((qxxq)/a+(qxxQ)/a+(qxxQ)/a)=0` or `Q=-q/2`

Discussion

147.	The capacitance of a parallel-plate air capacitor is 2 pF. If the air is replaced by a medium of dielectric constant 10. What will be its capacitance?
Answer» \(\cfrac{C_2}{C_1}\) = \(\cfrac{k_2}{k_1}\) C₂ = \(\cfrac{k_2}{k_1}\) x C₁ = \(\cfrac{10}1\) x 2 = 20 pF is the required capacitance.

147.

The capacitance of a parallel-plate air capacitor is 2 pF. If the air is replaced by a medium of dielectric constant 10. What will be its capacitance?

Answer»

\(\cfrac{C_2}{C_1}\) = \(\cfrac{k_2}{k_1}\)

C₂ = \(\cfrac{k_2}{k_1}\) x C₁ = \(\cfrac{10}1\) x 2 = 20 pF

is the required capacitance.

Discussion

148.	Three capacitors have capacities 2 µF, 4 µF and 8 µF. Find the equivalent capacity when they are connected in (a) series (b) parallel.
Answer» Data : C₁= 2 µF, C₂ = 4 µF, C₃= 8 µF (a) Series arrangement: The equivalent capacity, ∴ C_S = = 1.143 µF (b) Parallel arrangement: The equivalent capacity is C_P = C₁+ C₂+ C₃ = 2 + 4 + 8 = 14 µF

148.

Three capacitors have capacities 2 µF, 4 µF and 8 µF. Find the equivalent capacity when they are connected in (a) series (b) parallel.

Answer»

Data : C₁= 2 µF, C₂ = 4 µF, C₃= 8 µF

(a) Series arrangement:

The equivalent capacity,

∴ C_S = = 1.143 µF

(b) Parallel arrangement:

The equivalent capacity is

C_P = C₁+ C₂+ C₃ = 2 + 4 + 8 = 14 µF

Discussion

149.	When a capacitor having capacitance `4muF` and potential difference 100 volt is discharged, the energy released in joules isA. 0.01B. 0.02C. 0.03D. 0.07
Answer» Correct Answer - B `E=1/2CV^(2)=1/2xx4xx10^(-6)xx10^(4)=0.02J`

Discussion

150.	If 100 J of work must be done to move electric charge equal tp 4C from a place where potential is `-10V` to another place where potential si V volt, find the value of V.
Answer» Correct Answer - 15V Here, `W_(AB) = 100 J, q = 4C`, `V_(A) = -10 V, V_(B) = V = ?` As `W_(AB) = q(V_(B) - V_(A))` `:. 100 = 4 (V + 10), :. V = 15 vol t`.

Discussion

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