InterviewSolution
Saved Bookmarks
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.
| 1351. |
Define dielectric strength of a medium. What is its value for vacuum. |
| Answer» The maximum electric field that a dielectric medium can withstand without breaking down of its insulating property is called dielectric strength. Its value for vacumm is infinity. | |
| 1352. |
What is the effect of presence of a dielectric medium on (i) capacitance of a parallel plate capacitor (ii) electrostatic force between two charges ? |
|
Answer» With presence of dielectric medium, (i) Capacitance fo capacitor increases (ii) Electrostatic force between two charges decreses. |
|
| 1353. |
If dielectric strength of air (minimum field required for ionisation of a medium) is `3 M V//m`, can a metal sphere of radius 1cm hold a charge of 1 coulomb ? |
|
Answer» Here, `q = 1C and r = 1cm = 10^(-2) m`. Electric field intensity at the surface of the sphere, `E = (q)/(4pi in_(0) r^(2)) = (9xx10^(9)xx1)/((10^(-2))^(2))` As this value of E is much greater than the dielectric strength of air `(3 MV// = 3xx10^(6) V//m)`, therefore air around the sphere will get ionised and the entire charge from the sphere will leak out into air. Hence the sphere will leak out into air Hence the sphere of radius 1cm cannot hold 1 coulomb of charge in air. |
|
| 1354. |
Equipotential spheres are drawn round a point charge. As we move away from charge will the spacing between two spheres having a constant potential difference decrease, increase or remain constant. |
|
Answer» `V_1gtV_2` `V_1=1/(4piepsilon_0).q/r_1` and `V_2=1/(4piepsilon_0)` and `V_2=1/(4piepsilon_0).q/r_2` `Now, `V_1-V_2=q/(4piepsilon_0)(1/r_1-1/r_2)=q/(4piepsilon_0) ((r_2-r_1)/(r_1r_2)` `:. (r_2-r_1)=((4piepsilon_(0))(V_1-V_2))/q(r_1r_2)` For a constant potential difference `(V_1-V_2)` `r_2-r_1 prop r_1r_2` i.e. the spacing between two spheres `(r_2-r_1)` increases as we move away from the charge, because the product `r_1r_2` will increase. |
|
| 1355. |
(a). Calculation the largest possible electrostatic energy in `1cm^(3)` volume of air.t he dielectric breakdown of air happens when field exceeds `3xx10^(6)V//m` (b). A conducting ball of radius 10 cm is placed in distilled water `(in_(r)=80)` and charged with a charge `Q=2xx10^(-9)C`. Calculated the energy used up in charging the ball. |
|
Answer» Correct Answer - (a) `39.6muJ` (b). `2.25xx10^(-9)J` |
|
| 1356. |
The work done to move a charge along an equipotential from `A` to `B`A. cannot be defined as `- int_(A)^(B) E. dl`B. must be defined as `- int_(A)^(B) E. dl`C. is zeroD. can have a non-zero value |
|
Answer» Correct Answer - B::C From the knowledge of theorry, work done to move a charge along an equipotential from `A` to `B`, `W = - int_(A)^(B) vec(E). Vec(dl)` On equipotenital surface, `vec(E) _|_ vec(dl)` `W = - int_(A)^(B) E(dl) cos 90^(@) = Zero`. Choices (b) and (C ) are correct. |
|
| 1357. |
Assertion. Work done in moving any charge through any distance on an equipotential surface is zero. Reason. An equipotential surface is very smooth.A. both, Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. both, Assertion and Reason are true, but Reason is not the correct explanation of the Asserrtion.C. Assertion is true, but the Reason is false.D. both, Assertion and Reason are false. |
|
Answer» Correct Answer - c Assertion is true but the reason is false. |
|
| 1358. |
Assertion: Electrons move away from a low potential to high potential region. Reason: Because electrons have negative chargesA. both, Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. both, Assertion and Reason are true, but Reason is not the correct explanation of the Asserrtion.C. Assertion is true, but the Reason is false.D. both, Assertion and Reason are false. |
|
Answer» Correct Answer - a As electron has a negative charge. That is why it moves from a region of lower potential to a region of higher potential. |
|
| 1359. |
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the center of the cube.A. `(2q)/(sqrt(3) pi in_(0) b)` ,zeroB. `(4q)/(sqrt(3) pi in_(0) b)` ,zeroC. `(5q)/(sqrt(3) pi in_(0) b)` ,zeroD. `(7q)/(sqrt(3) pi in_(0) b)` ,zero |
|
Answer» Correct Answer - 2 `V_(0) = (1)/( 4 pi in_(0)) .(q)/(sqrt(3) b//2) xx 8 = (4)/(sqrt(3)) (q)/(pi in_(0) b)` `E_(o) = 0` |
|
| 1360. |
The equation of an equipotential line in an electric field is ` y = 2 x `, then electric field strength vectro at `1, 2` may be .A. ` 4 hat I + 3 hat j`B. ` 4 hat I + 8 hat j`C. ` 8 hat I + 4 hat I + 4 hat j`D. ` - 8 hat I + 4 hat j` |
|
Answer» Correct Answer - D Electric field will be perpendicyalr line ` y= 2x ` so it will have slope ` m =- 1/2`. |
|
| 1361. |
In a regular polgyon of n sides, each cormer is at a distance r from the center . Identical chargs are placed at ( n-1) corners . At the center, the intensity is E and the potential is `VM`. The ratio ` V//E` has magnitude .A. ` r n`B. ` r (n -1)`C. ` (n -1) //r`D. ` r(n -1) //n` |
|
Answer» Correct Answer - B `E =K q/r^2` and ` V= (K(n-1)q)/r` so ` V/E = ((n-1))/r xx r^2 rArr V/E = (n-1)r` |
|
| 1362. |
In a uniform electric fieled, the potential is ` 10V` at the origin of coordinates , and ` 8 V` at each of the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). The potential at the point (1, 1,1 ) will be .A. ` 0`B. ` 4 V`C. ` v`D. ` 1 0V` |
|
Answer» Correct Answer - B ` vec E = 2 hat I + 2 har j + 2 hat k` ` dV =- vec E . Vec (dr)` ` V-10 =- (D_x dx+E_y dy + E_z dz)` ` V= 10 =- (2 +2 +2) rArr V =4` volt . |
|
| 1363. |
The diagram show a small bead of mass `m` carrying charge `q`. The bead can freely move on the smooth fixed ring placed on a smooth borizontal plane . In the same pane a charge ` +Q` has also been fixed as shown. The potential at the point ` P` due to ` +Q` is `V` . The velocity which the bead should projected from the point `P` so that it can complete a circle should be greater than . A. ` sqrt ( (6qV)/(m))`B. `sqrt ((qV)/(m))`C. ` sqrt( (3qV)/(m))`D. none |
|
Answer» Correct Answer - A From energy conservation ` QV + 1/2 mu^2 =Q. 4 V + 0 rArr 1/2 mu^2 = 3 QV` ` rArr u = sqrt ( 6 QV)/m` . |
|
| 1364. |
Charges `+q` and `-q` are located at the corners of a cube of side as show in the figure. Find the work done to separate the charges to infinite distance |
|
Answer» Correct Answer - A::B::C::D `W_(external)=DeltaPE=(1)/(4piepsilon_0)(q^2)/(a)[-3/1+3/sqrt2-1/sqrt3]xx4` `=(1)/(4piepsilon_0)(q^2)/(a)*(4)/(sqrt6)[3sqrt3-3sqrt6-sqrt2]` |
|
| 1365. |
A point charge `+Q` is placed at a distance r from a short dipole of dipole moment `vecp` as shown in figure. The line joining the point charge Q to the centre dipole is perpendicular to dipole moment `vecp`, which of the following statement are true?A. The net electrostatic force on the dipole is zeroB. The electrostatic force on the dipole is `(Qp)/(4 pi epsi_(0)r^(3))` and is acting along the dipole moment pC. Torque acting on the dipole is `(Qp)/(4 pi epsi_(0) r^(2))` in the inward directionD. Torque acting on the dipole is `(Qp)/(2 pi epsi_(0)r^(2))` in the outward direction |
|
Answer» Correct Answer - B::C |
|
| 1366. |
A point charge q has been placed at a distance x from the centre of a neutral solid conducting sphere of radius R(x gt R). Find the potential of the sphere. How will your answer change if the sphere is not solid, rather it is a thin shell of conductor. |
|
Answer» Correct Answer - `(1)/(4pi in_(0))(q)/(x)` for both cases |
|
| 1367. |
A parallel plate capacitor of capacitance 100 μF is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes(a) 250 V (b) 500 V (c) 1000V (d) 2000 V |
|
Answer» (a) 250 V Here, C’ = 2C, since the charge remains the same. q = C’V’ = CV ⇒ V = \(\frac {CV}{2C}\,= \frac{500}{2}\) = 250V |
|
| 1368. |
A metallic solid sphere is placed in a uniform electric fied. The lines of force follow the path(s) shown in Figure as A. (a) 1B. (b) 2C. (c) 3D. (d) 4 |
|
Answer» Correct Answer - D The electric lines of force cannot enter the metallic sphere as electric field inside the solid metallic sphere is zero. Also, the origination and termination of the electric lines of force from the metallic surface is normally (directed towards the centre). |
|
| 1369. |
Two parallel infinite line charges with linear charge densities `+lambda C//m` and `-lambda` C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?A. zeroB. `(2 lambda)/(pi epsi_(0)R)`C. `(lambda)/(pi epsi_(0)R)`D. `(1)/(2 pi epsi_(0)R)` |
|
Answer» Correct Answer - B |
|
| 1370. |
Three charges Q, `+q` and `+q` are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to A. (a) `(-q)/(1+sqrt2)`B. (b) `(-2q)/(2+sqrt2)`C. (c) `-2q`D. (d)`+q` |
|
Answer» Correct Answer - B Here we have `(Qq)/(a)+(q^2)/(a)+(Qq)/(asqrt2)=0` `:.` `Q=-(qsqrt2)/(sqrt2+1)=-(2q)/(2+sqrt2)` |
|
| 1371. |
Two points P and Q are maintained at the potentials of 10V and -4V, respectively. The work done in moving 100 electrons from P to Q is:A. ` - 9 . 60 xx 10^(-17) J`B. ` - 9 . 60 xx 10^(-17) J`C. ` -2 . 24 xx 10^(-16) J`D. ` -2 . 24 xx 10^(-16) J` |
|
Answer» Correct Answer - D `W_(p rarr Q) |_(ext) =q (V_Q - V_p)` `=- 1. 6 xx 10^(19) xx 100 (-4 - 10)` `= 2 . 24 xx 10^(16) J`. |
|
| 1372. |
Two points P and Q are maintained at the potentials of 10V and -4V, respectively. The work done in moving 100 electrons from P to Q is:A. (a) `9.60xx10^-17J`B. (b) `-2.24xx10^-16J`C. (c) `2.24xx10^-16J`D. (d) `-9.60xx10^-17J` |
|
Answer» Correct Answer - C `(W_(PQ))/(q)=(V_Q-V_P)` `impliesW_(PQ)=q(V_Q-V_P)` `=(-100xx1.6xx10^-19)(-4-10)` `=+2.24xx10^16J` |
|
| 1373. |
Charges `q` is uniformly distributed over a thin half ring of radius `R`. The electric field at the centre of the ring isA. `(q)/(2 pi^(2) epsi_(0)R^(2))`B. `(q)/(4 pi^(2) epsi_(0)R^(2))`C. `(q)/(4 pi rpsi_(0)R^(2))`D. `(q)/(2 pi epsi_(0)R^(2))` |
|
Answer» Correct Answer - A |
|
| 1374. |
Energy of a capacitor of capacitance C, when subjected to a potential V, is given by (a) \(\frac{1}{2}\)CV2(b) \(\frac{1}{2}\) C2 V (c) \(\frac{1}{2}\) CV(d) \(\frac{1}{2}\) \(\frac {C}{V}\) |
|
Answer» Correct answer is (a) \(\frac{1}{2}\)CV2 |
|
| 1375. |
A circuit is shown in the figure below. Find out the charge of the condenser having capacity `5muF` A. `4.5muC`B. `9muC`C. `7muC`D. `30muC` |
|
Answer» Correct Answer - B From figure, charge on condenser of capacity `3muF` is, `Q_(1)=C_(1)V=3xx10^(-6)xx6=18xx10^(-6)C`. Thus, p.d. across `C_(1)` is, `V_(1)=(VC_(p))/(C_(1)+C_(p))" ""where"" "C_(p)=2+5=7muF` `=(7xx6)/10=4.2V` Thus, p.d. across `C_(p)` is, 1.8 V Now, charge on `5muF` condenser is, `Q_(2)=5xx10^(-6)xx1.8=9muC` |
|
| 1376. |
Two identical capacitors haveng plate separation `d_(0)` are connected parallel to each other across the points A and B as shown in. A charge Q is imparted to the system by removed Now the first plate of the first capacitor and the second plate of the second capacitor start moving with constant velocity `u_(0)` toward left. Find the magnitude of the current flouing in the loop during this process. . |
|
Answer» Correct Answer - 2 |
|
| 1377. |
A point charge q is placed inside a conducting spherical shell of inner radius 2R and outer radius 3R at a distance of R fro the centre of the shell. The electric potential at the centre of shell will (potential at infinity is zero).A. `(q)/(2R)`B. `(4q)/(3R)`C. `(5q)/(6R)`D. `(2q)/(3R)` |
|
Answer» Correct Answer - C |
|
| 1378. |
A `4 muF` capacitor, a resistance of `2.5 M Omega` is in series with `12V` battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [ Given In (2) = 0.693]A. 13.86 sB. 6.93 sC. 7 sD. 14 s |
|
Answer» Correct Answer - A |
|
| 1379. |
Two condensers of capacity `C_(1)` and `C_(2)`, are connected in series to a battery as shown in Fig. The adjoining graph shows the variraton of potential in going from `a` to `b`. Thereofore. A. `C_(1) gt C_(2)`B. `C_(1)= C_(2)`C. `C_(1) lt C_(2)`D. Cannot say |
|
Answer» Correct Answer - C Refer to Fig. Potential of outer plate of `C_(1)` is zero as negative of battery is earthed. As inner plate of `C_(1)` si at positive pot. The potential rises, Inner plate of `C_(2)` is at the same potential as inner plate of `C_(1)`. Therefore, potential remains constant. The graph shows that potential rises further. This indicates that `V_(2) gt V_(1)`. As `Q` is same in series combination, therefore `C_(2) gt C_(1)` or `C_(1) lt C_(2)`. |
|
| 1380. |
Two capacitors, one of `4muF` and the other of `5muF`, are connected in parallel and then charged on a 100 V supply. The charge on each capacitor is respectivelyA. 0.4 mC and 0.5 mCB. 0.4 mC and 0.8 mCC. 4 mC and 5 mCD. 4 mC and 8 mC |
|
Answer» Correct Answer - A `Q_(1)=C_(1)V=4xx10^(-6)xx100=0.4mC` `Q_(2)=C_(2)V=5xx10^(-6)xx100=0.5mC` |
|
| 1381. |
Four condensers each of capacity `4muF` are connected as shown in the adjoining figure. If `V_(P)-V_(Q)=15V`, the energy stored in the system is A. 2.4 ergsB. 1.8 ergsC. 3.6 ergsD. 5.4 ergs |
|
Answer» Correct Answer - B From figure, the equivalent capacity of the combination is, `1/C_(s)=1/4+1/((4+4))+1/4=5/8` `therefore" "C_(s)=8/5mmuF" ""Thus",` `E=1/2C_(s)V^(2)=1/2xx8/5xx10^(-9)xx225` `E=1.8xx10^(-7)J=1.8erg.` |
|
| 1382. |
A `4muF` capacitor is charged to 400 V. If its plates are joined through a resistance of `2kOmega`, then heat produced in the resistance isA. 0.16 JB. 0.32 JC. 0.64 JD. 1.28 J |
|
Answer» Correct Answer - B `C=4muF,V=400V,R=2kOmega,H=?` `H=E=1/2CV^(2)` `=(4xx10^(-6)xx16xx10^(4))/2=64/2xx10^(-2)=0.32J` |
|
| 1383. |
If a `4muF` capacitor is charged to 1 kV, then energy stored in the capacitor isA. 1 JB. 4 JC. 6 JD. 2 J |
|
Answer» Correct Answer - D `C=4muF,V=1kV,E=?` `E=1/2CV^(2)=(4xx10^(-6)xx10^(6))/2=2J` |
|
| 1384. |
In the circuit shown in figure 3.141, find the energy stored in `4muF` capacitor in steady state. A. `20 muC`B. `40 muC`C. `10 muC`D. `120 muC` |
|
Answer» Correct Answer - A |
|
| 1385. |
A capacitor of capacitance `10 mu F` is charged to a potential `50 V` with a battery. The battery is now disconnected and an additional charge `200 mu C` is given to the positive plate of the capacitor. The potential difference across the capacitor will be.A. 50 VB. 80 VC. 100 VD. 60 V |
|
Answer» Correct Answer - D |
|
| 1386. |
Conisder a thin spherical shell of radius R with centre at the origin, carrying uniform poistive surface charge denisty. The variation of the magnitude of the electric field `|vecE(r)|` and the electric potential V(r) with the distance r from the centre, is best represented by which graph?A. B. C. D. |
| Answer» Correct Answer - C::D | |
| 1387. |
Conisder an electric field `vecE=E_0hatx` where `E_0` is a constant . The flux through the shaded area (as shown in the figure) due to this field is A. ` 2 E_0a^2`B. ` sqrt 2 E_0 a^2`C. ` E_0 a^2`D. ` (E_0a^2)/(sqrt 2)` |
|
Answer» Correct Answer - C ` pi = int overline . Overline(dS) =` Ex projected area perpendicualr of e `(x - axis ) = E xx a^2)`. |
|
| 1388. |
Which of the field patterns given below is valid for electric field as well as for magnetic field?A. (a) B. (b) C. (c) D. (d) |
|
Answer» Correct Answer - C The pattern of field lines shown in option (c) is correct because (a) a current carrying toroid produces magnetic field lines of such pattern (b) a changing magnetic field with respect to time in a region perpendicular to the paper produces induced electric field lines of such pattern. |
|
| 1389. |
Conisder an electric field `vecE=E_0hatx` where `E_0` is a constant . The flux through the shaded area (as shown in the figure) due to this field is A. `2E_(0) a^(2)`B. `sqrt(2) E_(0) a^(2)`C. `E_(0)a^(2)`D. `(E_(0)a^(2))/sqrt(2)` |
| Answer» Correct Answer - C | |
| 1390. |
Which of the following statement(s) is/are correct ?A. If the electric field due to a point charge varies as `r^(-2.5)` instead of `r^(-2)`, then the Gauss law will still be valid.B. The Gauss law can be used to calculate the field distribution around an electric dipole.C. If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.D. The work done by the external force in moving a unit positive charge from point a at potential `V_(A)` to point B at potential `V_(B)` is `(V_(B)-V_(A))`. |
| Answer» Correct Answer - C | |
| 1391. |
A wooden block performs SHM on a frictionless surface with frequency, `v_(0)`. The block carries a charge `+Q` on its surface. If now a uniform electric field `vec (E)` is switched on as shwon in figure., then the SHM of the block will be A. of the same frequency and with shifted mean position.B. of the same frequency and with the same mean position.C. of charged frequency and with shifted mean position.D. of Charged frequency and with the same mean position. |
| Answer» Correct Answer - A | |
| 1392. |
An electric dipole has the magnitude of its charge as `q` and its dipole moment is `p`. It is placed in a uniform electric field `E`. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectivelyA. `2q.E` and minimumB. `q.E` and `p.E`C. zero and minimumD. `q.E` and maximum |
|
Answer» Correct Answer - 3 `tau = pE sin 0^(@) = 0` `U = -pE cos 0^(@) = -pE = U_(min)` |
|
| 1393. |
An electric dipole has the magnitude of its charge as `q` and its dipole moment is `p`. It is placed in a uniform electric field `E`. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectivelyA. `2qE` and minimumB. `qE` and `pE`C. zero and minimumD. qE and maximum |
|
Answer» Correct Answer - C |
|
| 1394. |
A bullet of mass `2 gm` is having a charge of `2 muc`. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of `10 m//s`A. `5kV`B. `50 kV`C. `5V`D. `50V` |
|
Answer» Correct Answer - B |
|
| 1395. |
Four electric charges `+q, +q, -q` and `-q` are placed at the corners of a square of side 2L (see figure). The electric potential at point `A`, mid-way between the two charges `+q` and `+q`, is A. `(1)/(4pi epsilon_(0)) (2q)/(L) (1+(1)/(sqrt(5)))`B. `(1)/(4pi epsilon_(0)) (2q)/(L) (1-(1)/(sqrt(5))`C. zeroD. `(1)/(4pi epsilon_(0))(2q)/(L) (1+sqrt(5))` |
|
Answer» Correct Answer - B |
|
| 1396. |
Two identical charges `+Q` are kept fixed some distance apart.A small particles `P` with charge `q` is placed midway between them . If `P` is given a small displacement `Delta` ,it will undergo simple harmonic motion if (i) `q` is positive and `Delta` is along the line joining the charges (ii) `q` is positive and `Delta` is perpendicular to the line joining the charges (iii) `q` is negative and `Delta` is perpendicular to the line joining the charges (iv) `q` is positive and `Delta` is along the line joining the chargesA. `q` is positive and `Delta` is along the line joining the charges.B. `q` is negative and `Delta` is perpendicular to the line joining the charges.C. `q` is negative and `Delta` is perpendicular to the line joining the charges.D. `q` is positive and `Delta` is along the line joining the charges. |
|
Answer» Correct Answer - B::C |
|
| 1397. |
Two identical charges `+Q` are kept fixed some distance apart.A small particles `P` with charge `q` is placed midway between them . If `P` is given a small displacement `Delta` ,it will undergo simple harmonic motion if (i) `q` is positive and `Delta` is along the line joining the charges (ii) `q` is positive and `Delta` is perpendicular to the line joining the charges (iii) `q` is negative and `Delta` is perpendicular to the line joining the charges (iv) `q` is positive and `Delta` is along the line joining the chargesA. `(i) ,(ii)`B. `(i),(iii)`C. `(ii),(iii)`D. `(i),(iv)` |
| Answer» Correct Answer - 2 | |
| 1398. |
Find the `V_(ab)` in an electric field `E=(2hati+3hatj+4hatk)N/C`, where `r_a=(hati-2hatj+hatk)m` and `r_b=(2hati+hatj-2hatk)m` |
|
Answer» Correct Answer - A Here the given field is uniform (constant) so Using `dV=-E.dx` `or V_(ab)=V_a-V_b=-int_b^aE.dr` `=-int_((2,1,-2))^((1,-2,1))(2hati+3hatj+4hatk).(dxhati+dyhatj+dzhatk)` `-int_((2,1,-2))^((1,-2,1))(2 dx+3dy+4dz)` `-[2x+3y+4z]_((2,1,-2))^((1,-2,1))` `=-1V` |
|
| 1399. |
The electric field at a distance 2 cm from the centre of a hollow sphereical coducting shell of radus 4 cm having a charge of `2xx10^C` on its surface isA. `1.1xx10^10V//m`B. `4.5xx10^I-10V//m`C. `4.5xx10^10V//m`D. zero |
|
Answer» Correct Answer - D E=0 inside a hallow charged spherical conducting shell. |
|
| 1400. |
Charge Q is given a displacement `r=ahati+bhatj` in electric field `E=E_1hati+E_2hatj`. The work done isA. `Q(E_1a+E_2b)`B. `Qsqrt((E_1a)^2+(E_2b)^2)`C. `Q(E_1+E_2)sqrt(a^2+b^2)`D. `Qsqrt(E_1^2+E_2^2) sqrt(a^2+b^2)` |
|
Answer» Correct Answer - A `W=F.r` `=(QE).r=Q(E.r)` ltbr `=Q(E_1a+E_2b)` |
|