Given below are two statements : Statement I : When an electron is brought near the surface of a metal in any dielectric medium, an image charge of opposite polarity is included inside the metal at the same distance from the surface as that of the electron in the dielectric medium. Statement II : In the case of metal-semiconductor junction, an electron in the semiconductor also creates an image charge in the metal which decreases the energy barrier height. In the light of the above statements, choose the correct answer from the options given below :
Both Statement I and Statement II are true
Statement I — the method of images. TRUE. A charge placed near a conducting surface induces a surface-charge distribution on the metal. Electrostatically, the field in the region outside the metal is identical to that produced by removing the conductor and placing an equal and opposite image charge symmetrically behind the surface — at the same perpendicular distance on the other side. So an electron a distance x from the metal behaves as though a charge +q sits at −x, and the attractive force is
\(F=-\dfrac{q^{2}}{4\pi\varepsilon(2x)^{2}}=-\dfrac{q^{2}}{16\pi\varepsilon x^{2}}\)
The permittivity ε of the surrounding dielectric appears in the force, which is why the statement mentions the dielectric medium.
Statement II — image-force barrier lowering at a Schottky junction. TRUE. The same image attraction acts on an electron in the semiconductor as it approaches the metal. Integrating the force gives an image potential energy \(-\dfrac{q^{2}}{16\pi\varepsilon_s x}\), and adding it to the potential energy of the depletion-region field \(-qEx\) produces a barrier whose peak is pulled down and shifted away from the interface. The reduction is
\(\Delta\phi_B=\sqrt{\dfrac{qE}{4\pi\varepsilon_s}}\)
This is Schottky barrier lowering, and its consequences are real and measurable: the reverse current of a Schottky diode is not perfectly saturated but grows with reverse bias, because increasing E lowers the barrier further. The effect also increases with electric field, so it matters most in heavily doped, high-field junctions.
Relationship between the two. Statement II is the direct application of Statement I to the metal–semiconductor contact, so not only are both true, the second follows from the first.
Hence, both Statement I and Statement II are true.
For a uniform line charge of 8 nc/m lying along the z axis, the electric field at a radius of 3 m from the uniform line is given by :
Match the following lists in terms of relative permittivity :
| List – I | List – II |
| a. Air | i. 25 |
| b. Alcohol (Ethyl) | ii. 4 to 7 |
| c. Glass | iii. 80 |
| d. Fresh water | iv. 1 |
Three point charges q are placed at the corners of an equilateral triangle. Another point charge −Q is placed at the centroid of the triangle. If the force on each of the charges q vanishes, then the ratio Q/q is
The components of the electric field, in a region of space devoid of any charge or current sources, are given to be E i= a i+ Σ j=1,2,3 bij xj , where a iand b ij are constants independent of the coordinates. The number of independent components of the matrix b ij , is
Whenever a conductor cuts magnetic flux, an e.m.f. is induced in that conductor. This phenomenon is according to
The value of electric field E at a point in Electric field of a point charge can be calculated using:
An inductor of 3.3mH with a series resistance of 12.5 ohms is connected to a 5V dc supply. When the supply is switched off, the circuit current decay to zero in 60 microseconds. What is the value of back e.m.f. generated?