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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.
| 38901. |
A solenoid of length 0.4 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 10 A. What is the magnitude of the magnetic field inside the solenoid ? |
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Answer» |
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| 38902. |
In a n-type semiconductor the donor energy level is slightly _____ |
| Answer» SOLUTION :below the BOTTOM END of CONDUCTION BAND | |
| 38903. |
An oscillator circuit consists of an inductance of 0.5 mH and capacitor of 20µF. The resonance frequency of the circuit is |
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Answer» 15.92 Hz |
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| 38904. |
In one dimensional motion ,instantaneous speed V satisfies 0leVltV_(0) |
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Answer» The displacement in time T must always take nonnegative values. |
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| 38905. |
White light reflectedat normal incidence from a soap film, has maximum at 6000 Åand the minimum at 4500 Å in the visible region with no minimum in between.If mu = 1.33 for the film, the thickness of the film is : |
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Answer» `0.23 MU m` For MAXIMA, `lambda_(1) = 6000 Å` For MINIMA, `lambda_(2) = 4500 Å` Applying the condition for maxima `2mu T = (2n -1).(lambda_(1))/(2)`....(1) and for minima `2mut = 2n .(lambda_(2))/(2)`...(2) From (1) and (2) `(2n - 1) .(lambda_(1))/(2) = n lambda_(2)` `(2n -1)6000 = 2n XX 4500` `therefore`n = 2 Now from eqn.(2) `2 xx 1.33 xx t = 2 xx 2xx(4500)/(2)` `therefore tapprox 0.34 mu m`. |
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| 38906. |
In a semiconductor, is the energy gap between the valence and conduction bands is 1.1 eV It is expressed in joules as |
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Answer» `1.2xx10^-19J` |
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| 38907. |
A drop of water of 0.2 g mass is placed between two well cleaned glass plates, the distance between which is 0.01 cm. Find the force of attraction between the plates, |
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Answer» Solution :Suppose the drop spreads evenly and as SEEN from above has the shape of a circle of RADIUS R (Fig. 21.9). The AREA of this circle is `S = V//d = m//rhod`. The FORCE of Fig. 21.7. attraction between the plates is `F = DeltapsS` where `Deltap= 2 sigma//d` is the excess pressure under the curved surface. Therefore `F=2sigma m// rho d^(2)` |
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| 38908. |
An object 1 cm tall is placed 4 cm infront of a mirror. In order to produce an upright image of 3 cm height one needs a |
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Answer» Convex mirror of RADIUS of CURVATURE 12cm |
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| 38909. |
If the relation between range R and time of flight T is given by R = 5 T2 , the angle of throw of the projectile is : |
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Answer» Solution :Here `(2v^(2)SINTHETACOSTHETA)/G=(5xx4v^(2)SIN^(2)theta)/g^(2)` or `tantheta=1` if `g=10ms^(-2) or theta=45^@` |
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| 38910. |
Consider a harmonic wave travelling on a string of mass per unit length p.The wave has a velocity v, amplitude A and frequency f The power transmitted by a harmonic wave on the string is proportional to (take constant of proportionality as 2pi^(2) |
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Answer» <P>`mu` `P = 2pi^(2)F^(2)A^(2)rhovs = 2pi^(2)f^(2)A^(2)mu v` |
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| 38911. |
A car travelling at a speed of 8 m/s towards a large wall horns a sound of frequency 130 Hz if the person stands behind the car such that the car receding from him approaches the wall the no. of beats heard by him per second is (velocity of speed in air 340 m/s) |
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| 38912. |
The De Broglie wavelength of a particle is equal to that of a photon, then the energy of photon will be |
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Answer» EQUAL to the KINETIC energy of the PARTICLE |
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| 38913. |
If a current carrying conductor is placed in the direction of the magnetic field , then the force on it is _______ . |
| Answer» SOLUTION :N/A | |
| 38914. |
The glandular tissue of breast is divided into |
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Answer» 10-12 MAMMARY lobes |
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| 38915. |
A photoelectric surface is illuminated successively by monochromatic light of wavelength lambda and (lambda)/(2).If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times in the first case,the work function of the surface of the material is: (h=Plank's constant ,c=speed of light) |
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Answer» `(HC)/(3lambda)` In first case ,`K_(1)=(hc)/(lambda)-phi` In second case ,`K_(2)=(hc)/((lambda)/(2))-phi=(2hc)/(lambda)-phi` But `K_(2)=3K_(1)` is given . `THEREFORE (2hc)/(lambda)-phi=(3hc)/(lambda)-3phi` `3phi-phi=(3hc)/(lambda)-(2hc)/(lambda)` `2phi=(hc)/(lambda)` `therefore phi=(hc)/(2lambda)` |
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| 38916. |
If the pressure of a gas is doubled keeping its volume and mass constant, the r.m.s. velocity of molecules of the gas will |
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Answer» Not change |
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| 38917. |
Which mirror is to be used to obtaina paralle beam of light from a small lamp? |
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Answer» a) PLANE mirror |
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| 38918. |
If the velocity of a car in increased by 20%, then the minimum distance in which it can be stopped increases by: |
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Answer» 0.44 or `S/S=(1+(20)/(100))^(2)=(36)/(25)` or `((S)/(S)-1)xx100=((36)/(25)-1)xx100`=44% |
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| 38919. |
A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Omega is connected to a 230 V variable frequency supply. (a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value. (b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power. (c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies? (d) What is the Q-factor of the given circuit? |
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Answer» Solution :(a) `omega_(0)=4167 "rad s"^(-1), v_(0)=663Hz` `I_(0)^("max")=14.1A` (b) `BARP=(1//2)I_(0)^(2)R` which is maximum at the same FREQUENCY (663 Hz) for which `I_(0)` is maximum `barP_("max")=(1//2)(I_("max"))^(2)R=2300W`. (c ) At `omega=omega_(0)PM Deltaw` [Approximation good if `(R//2L) lt lt omega_(0)`]. `Delta omega=R//2L=95.8"rad s"^(-1), Deltav=Delta omega//2pi =15.2Hz.` Power absorbed is half the peak power at n = 648 Hz and 678 Hz. At these FREQUENCIES, current amplitude is`(1//sqrt(2))` times `I_(0)^("max")`, i.e., current amplitude(at half the peak power points) is 10 A. (d) Q = 21.7 |
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| 38920. |
A point (sound generating) source of power 6.5 milli watts, is placed at the centre of the hollow cylinder of length 24cm and it.s radius of cross-section 5cm. Then the power passing through the lateral surface of the cylinder in milli watts is |
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| 38921. |
Monochromatic light of wavelength 6000 A travels through a glass of refractive index 3/2. The distance travelled by the wavefront in 1 picosecond is : |
| Answer» Answer :B | |
| 38922. |
A box contains a mixture of H_(2) and He gases. Which of the following statements are correct? |
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Answer» The average translational kinetic ENERGIES of `H_(2)` molecules and He atoms are same `E_("translational")=(3)/(2)RT` while total energy of diatomic molecule will be greater than that of monatonic. `E_("diatomic")=(5R)/(2),E_("monoatomic")=(3)/(2)RT` The average energy of `H_(2)` molecule is greater than that of He. |
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| 38923. |
A horizontal photoelectric plate whose work function is 1.9 eV is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically upward component of velocity of ejected photoelectrons is non-positive? |
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Answer» Solution :MINIMUM energy of the photon falling on photoelectric plate `=hv=(hc)/(LAMDA)=((6.625xx10^(-34))(3xx10^(8)))/(250xx10^(-19))` ACCORDING to Einstein's photoelectric equation, `(1)/(2)mv^(2)=(hc)/(lamda)-W` `[v_(min)]=(2)/(m)[(hc)/(lamda)-W]` `=(2)/(9.1xx10^(-31))[((6.625xx10^(-34))(3xx10^(8)))/(250xx10^(-9))-1.9xx(1.6xx10^(19))]` `=1.08xx10^(12)` `v_(min)=1.04xx10^(6)ms^(-1)` |
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| 38924. |
Formula of magnification of compound microscope is …........ . |
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Answer» `m=(LDf_e)/(f_0)` |
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| 38925. |
A point of object is placed at a distance 25 cm from a convex lens of focal length 20 cm. If a glassslab of thickness and refractive index 1.5 is inserted between the lens and object, image is formed at. Thickness t is found to be K times of 5 cm. Find K. |
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| 38926. |
A horizontal pipe line carries water in a stream line flow. At a point along the pipe where cross-sectional area is 10 cm^(2), the water velocity is 1 m/s and the pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is 5 cm^(2), is (density of water= 10^(3) kg.m^(-3)) : |
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Answer» 1000 Pa `A_(1)v_(1)=A_(2)v_(2)` `10xx1=5xxv_(2)rArrv_(2)=2`m/s Now by Bernoulli.s theorem, `(P_(1))/p+1/2v_(1)^(2)=(P_(2))/p+1/2v_(2)^(2)` `rArrP_(2)=P_(1)+1/2p(v_(1)^(2)-v_(2)^(2))` `=2000+1/2xx10^(3)(1^(2)-2^(2))` = 2000 - 1500 = 500 pascals. Thus CORRECT choice is (b). |
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| 38927. |
Find the mean free path of electrons in copper and compare it with the interatomic distance. |
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Answer» The INTERATOMIC distance is found from the concentration of atoms: `d=n_(A)^(-1//3)=2.3Å`. |
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| 38928. |
What is term modulation? |
| Answer» SOLUTION :Modulation is the PROCESS by which some characteristics, usually amplitude, frequency or PHASE of angle of high frequency, carrier wave is varied in ACCORDANCE with the instantaneous VALUE of the low frequency audio signal, called modulating signal. | |
| 38929. |
When in a coil current changes from 6a to 1 A in 0.05sec, the induced e.m.f. in it is 10V. The self-inductance of the coil is: |
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Answer» 0.1H |
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| 38930. |
Consider the following circuit with a switch S_(w). Potential difference of the cell is V. Initially, the switch is in position 1, connected to the left branch. After a long time the switch is shifted to position 2, to connect with the right branch. After the switch is shifted to position 2, find: (a) charge flowing through the switch. (b) work done by the battery. (c) heat loss in redistribution of charge. |
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Answer» Solution :We can easily understand that in both the states of switch equivalent capacitance (C ) remains same, hence total stored CHARGE (Q = CV) also remains same. In both the states parallel combination of `C_(1)` and `C_(2)` is connected in series with another CAPACITOR `C_(1)` . We can write C and Q as follows: `C=(C_(1)(C_(1)+C_(2)))/(2C_(1)+C_(2))""...(i)` `Q=CV rArr Q=(C_(1)(C_(1)+C_(2))V)/(2C_(1)+C_(2)) "" ...(ii)` In the network `C_(1) and C_(2)` are connected in parallel and total charge Q is divided between them, proportional to their capacitance. Hence, charge stored in `C_(1) and C_(2)`, connected in parallel, is `Q_(1)=(QC_(1))/(C_(1)+C_(2)) and Q_(2)=(QC_(2))/(C_(1)+C_(2))`respectively. Distributionof charge in both the states of the circuit is shown in the figures.  Fig-(a) shows charge distribution in state 1 of the circuit (when the switch is in position 1). Fig-(b) shows charge distribution in state 2 (when the switch is in positon 2) of the circuit. Fig-(c ) shows additional charge flowing through the various branches of circuit to adjust the charge distribution when the circuit changes its state. (a) LOWER plate of the middle capacitor has charge `(+QC_(2))/(C_(1)+C_(2))` in the first state and in the second state the same plate has to acquire final charge `(-QC_(2))/(C_(1)+C_(2))`. Hence, charge `(+2QC_(2))/(C_(1)+C_(2))` will be released by this plate, which will flow through the switch as shown in Fig-(c ). On substituting value of Q from equation (ii) we get charge flowing through the switch as follows: Charge through switch `=(+2(C_(1)(C_(1)+C_(2))V)/(2C_(1)+2C_(2))C_(2))/(C_(1)+C_(2))=(2C_(1)C_(2)V)/(2C_(1)+C_(2))` (b) Lower plate of the left capacitor`(C_(1))` has a charge `(+QC_(1))/(C_(1)+C_(2))` in the first state and the same plate has to acquire +Q charge in the second state of the circuit. Hence, charge `(+QC_(2))/(C_(1)+C_(2))` MUST flow from positive terminal of the batery towards the lower plate of left capacitor. Hence, additional charge `(+QC_(2))/(C_(1)+C_(2))` is supplied by the positive terminal battery when switch is shifted from position-1 to the positon-2. Work done by the battery = (Charge supplied by the battery) `xx` (potential difference of battery) `W_(b)=(QC_(2))/(C_(1)+C_(2))V=((C_(1)(C_(1)+C_(2))V)/(2C_(1)+2C_(2))C_(2))/(C_(1)+C_(2))V=(C_(1)C_(2)V^(2))/(2C_(1)+C_(2))` (c ) We already know that equivalent capacitance in both the states of the circuit is same and that potential difference applied is also the same. Hence, energy stored in the circuit has to be the same in both states. `U_(1)=U_(2)` According to energy conservation: `U_(1)+W_(b)=U_(2)+H` `rArr H=W_(b)=(C_(1)C_(2)V^(2))/(2C_(1)+C_(2))` |
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| 38931. |
Explain the scattering of light with an example. |
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Answer» Solution :Scattering of light is the process due to which light gets deflected and diffused all over as a result of its interplay with tiny matter PARTICLES like air/gas molecules etc. For scattering particles of EXTREMELY SMALL size the amount of scattering is inversely proportional to the fourth power of the wavelength of light. At sunset or sunrise, the Sun.s rays have to pass through a larger distance in the atmosphere and so most of the blue and other shorter wavelengths are removed by scattering. The least scattered light reaching our EYES is predominantly of red colour. Due to this, the Sun at sunset or sunrise appears to be REDDISH. |
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| 38932. |
(a) Define a wavefront. UsingHuygen'sprinciple, verify the laws of reflection at a plane surface . (b) In a singleslit diffraction experiment the wildth of the slit is made doublethe orignal width. Howdoes this affect the size and intensity of the central diffraction band ? Explain.(c) When a tiny circular obstaclein placedin thepath of lightfrom a distance source, a brightspot isseen at the centre of the obstacle . Explain why. |
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Answer» Solution :(a) Wavefront: Awavefront is define as the continous locus of all such particleof the medium whichare vibratingin same phase at any instant. There are different type of wavefront. (i) Spherical wavefront(ii)Plane wavefront(iii) Cylinderical wavefront. `BC = BC` `/_ BAC = /_BDC = 90^(@)` `AC = BD` `DeltaABC = DetlaADC ` `i = r` (b)In diffraction, `d sin theta = lambda` `dtheta = lambda` `theta = lambda //d` LINEAR width of CENTRAL maximun : If D is thedistanceof the screen from thesingleslit then linear width of central maximum will be. `beta_(0) = D xx 2theta = (2Dlambda)/(d)` If we increase the size with 2d the `beta_(0) = (2Dlambda)/(2d)` `2BETA = beta_(0) ""beta = (beta_(0))/(2)` Intensity of central maximum becomes 4 times. (c) When tinycircular obstacleis placed in the path of light from a distance source a bright spot is seen at the centre. lightwhen passes from broder of circularobstacle is diffractstowards centre ofthe obstacle in centre part all diffracted wavefront meet. That is why centre becomes bright.  
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| 38933. |
The resistance R of a wire is found by determining its length l and radius r. The percentage errors in measurement of l and r are respectively 1% and 2%. The percentage error in measurement of R is |
| Answer» ANSWER :B | |
| 38934. |
The radioactivity of a sample is R_1 at a time t_1 and R_2 at time t_2. If half life of sample is T, then no. of atoms that have disintegrated in time (t_2 - t_1) is proportional |
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Answer» `R_(1)t_(1)-R_(2)t_(2)` As `N_(2) lt N_(1)`, so no. of atoms decayed in TIME `(t_(2)-t_(1))` is `N_(1)-N_(2)=(R_(1))/(lambda)=(R_(1))/(lambda) -R_(2)/lambda =((R_(1)-R_(2))/(lambda))` `rArr N_(1)-N_(2)=((R_(1)-R_(2))T)/(0.693) ALPHA (R_(1)-R_(2))T` |
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| 38935. |
A perfectly white body |
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Answer» which REFLECTS all the wavelength INCIDENT on it |
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| 38936. |
Givethe applications of ICT in miningand agriculturesectors. |
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Answer» Solution :AgricultureThe implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the challenges and risk FACTORS. (a) ICT is widely used in increasing food productivity and farm management. (b) It helps to optimize the use of water, SEEDS and fertilizers etc. (c) Sophisticated technologies that include ROBOTS, temperature and moisture sensors, aerial images, and GPS technology can be used. (d) Geographic information systems are extensively used in farming to decide the SUITABLE place for the species to be planted. (iii) Mining (a) ICT in mining improves operational efficiency, remote monitoring and disaster LOCATING system. (b) Information and communication technology provides audio-visual warning to the trapped underground miners. (c) It helps to connect remote sites. |
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| 38937. |
A double convex thin lens made of glass ("refractive index" mu = 1.5) has both radii of curvature of magnititude 20 cm Incident light rays parallel to thte axis of the lens will converge at a distance L such that |
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Answer» L = 20 cm Give `R_(1) = + 20 cm, R_(2) = - 20 cm, mu = 1.5` `rArr = 20 cm`. Parallel rays converge at FOCUS. So L = f. |
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| 38938. |
The number of electric field lines emerged out from 1muC charge is ....... |
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Answer» `1.13 xx 10^(11)` `N =q/epsilon_(0) =10^(-3)/(8.85 xx 10^(-12))` `THEREFORE N = 0.11299 xx 10^(12.3) =1.13 xx 10^(8)` |
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| 38939. |
A rotary potentiometer has a circular resistance (C) and a conducting wiper (W) can rotate over it. An ideal battery and an ideal voltmeter are connected to it as shown in the figure. As the wiper is rotated on the circular resistance the reading of the voltmeter changes from zero to 20 V. What is reading of the voltmeter when wiper is at angular position theta = 120^(@) ? Assume that resistance C is almost a complete circle and resistance of all connecting wires and knobs in the circuit is zero. |
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Answer» |
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| 38940. |
Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, A_(v), of the amplifier is given by A_(v)= -(beta_(ac) R_(I))/(r_(i)), where beta_(ac) is the current gain, R_(L) is the load resistance and r_(i ) is the input resistance of the transistor. What is the significance of the negative sign in the expression of the voltage gain ? |
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Answer» Solution :`V_(c c)=V_(cg)+I_(c )R_(z)` Likewise, the input loop gives `V_(B B)=V_(B B)+I_(B)R_(B)` Where `v_(i)` is not zero, we get `V_(B B)+V_(i)=V_(B B)+I_(B)R_(B)+Delta I_(B)(R_(B)+r)` The change in `V_(Bg)` can be related to the input resistance `r_(i)` and the change in `I_(g)`. Hence, `v_(i)=Delta I_(B)(R_(B)+r_(i))` `v_(i)=r Delta I_(B)` The change in `I_(B)` causes a change in `I_(C )` . We define a parameter `beta_(ax)`, which is similar to the `beta_(dc)`  `beta_(ac)=(Delta I_(C ))/(Delta I_(B))=(i_(c ))/(i_(b))` which is also known as the ac current gain `A_(C )`. Usually `beta_(ac)` is dose to `beta_(dc)` in the linear region of the output characteristics. The change in `I_(c )` due to a change in `I_(B)` causes a change in `V_(CE)` and the voltage DROP across the resistor `R_(L)` because `V_(C C)` is FIXED. These CHANGES can be given by `Delta V_(C C)=Delta V_(CE)+R_(L) DeltaI_(C )=O` or `DeltaV_(CE)= -R_(L)Delta IC` The change in `V_(CB)` is the output voltage `v_(O)`. We get `v_(O)=Delta V_(CE)= - beta_(ac)R_(L)Delta I_(B)` The voltage gain of the AMPLIFIER is `A_(v)=(v_(o))/(v_(i))=(Delta V_(CB))/(r Delta I_(B)) implies A_(v)= - (beta_(ac)R_(L))/(r )` The negative sign indicates that output is phase reversed by an ANGLE `180^(@)`. |
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| 38941. |
A magnet is brought towards a coil (i) speedily (ii) Slowly, then the induced e.m.f/induced charge will be respectively |
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Answer» More in FIRST case/ More in first case |
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| 38942. |
Two identical bar magnets each of dipolemoment M and length l are perpendicular to each other themagnetic dipole moment of the combinationis |
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Answer» `2M` |
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| 38943. |
Parthenium or carrot grass has become ubiquitous in occurrence and causes pollen allergy. Parthenium came into India as a contaminant with imported |
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Answer» Wheat |
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| 38944. |
A wave is represented by equation y = 0. 01 sinpi(01 x + 3t) meter. The wave velocity is |
| Answer» Answer :C | |
| 38945. |
What should be the velocity of anelectron so that its momentum becomes equal to that of a photon of wavelength 5200 A^@ ? |
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Answer» `10 ^3 m s ^(-1)` |
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| 38946. |
A particle of mass m attached to the end of string of length l is released from the horizontal position. The particle rotates in a circle about O as shown When it is vertically below O, the string makes contact with a nail N placed directly below O at a distance h and rotates around it. For the particle to swing completely around the nail in a circle. |
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Answer» `h lt(3)/(5)L` |
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| 38947. |
A simple pendulum of length 1 m has a bob of 200 g suspended to a fixed point. It is displaced through 60^(@) and then released. What is K.E. when its inclination is 30^(@) with the vertical ? (g = 10 ms^(-2)) |
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Answer» 0.73 J  Similarly,`h-l-l cos 30^(@)` `:.` K.E. = LOSS of P.E. =mgh-mgh. =`MG(h-h.)=mg[l-l cos 60^(@)-l+l cos 30^(@)]` =`mgl[cos 30^(@)-cos 60^(@)]` =`0.2xx10xx1[(1.732-1)/(2)]0.732 J` |
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| 38948. |
The currents flowing in the two coils of self-inductance L_(1)=16 mH and L_(2)=12 mH are increasing at the same rate. If the power supplied to the two coils are equal, find the ratio of (i) induced voltages, (ii) the currents and (iii) the energies stored in the two coils at a given instant. |
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Answer» <P> Solution :(i) Induced emf (voltage) in a coil `epsilon = - L (dI)/(dt)``(epsilon_(1))/(epsilon_(2)) = (L_(1) (dI_(1))/(dt))/(L_(2) (dI_(2))/(dt)) = (L_(1))/(L_(2)) = 4/3` (ii) POWER supplied, `P = eI` As power is same for both coils `epsilon_(1)I_(1) = epsilon_(2)I_(2)` `rArr (I_(1))/(I_(2)) = (epsilon_(2))/(epsilon_(1))=3/4` (III) Energy stored in a coil, `E = 1/2 LI^(2)` `:. (E_(1))/(E_(2)) = (I/2 L_(1)I_(1)^(2))/(I/2 L_(2)I_(2)^(2)) =(L_(1)I_(1)^(2))/(L_(2)I_(2)^(2))=3/4` |
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| 38949. |
An electron moves with a constant speed v along a circle of radius r. Its magnetic moment will be ______ . (e is the electron's charge.) |
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Answer» evr `A=pir^(2)` Magnetic moment m = NIA `thereforem=I(pir^(2))` An ELECTRON moves with a CONSTANT SPEED v along a circle of radius r. The current produced, `I=(ev)/(2pir)` `thereforem=(ev)/(2pir)xxpir^(2)` `thereforem=1/2evr` |
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| 38950. |
Two infinitely long. Parallel, thin sheets, having surface charge densities + sigma and -sigmarespectivelyare separated. By a small distance. The medium between the sheets is vacuum. The electric field in the region between the sheets is |
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Answer» ZERO |
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