Figure represents a crystal unit of cesium chloride, CsCl. The cesium atohis, represented by open circles are situated at the comers of a cube of side 0.40 nm, whereas a Cl atom is situated at the centre of the Cube.The Cs atoms are deficient in one electron while the Cl atom carries an excess electron.
(i) What is the net electric field on the Cl atom due to eight Cs atoms? (ii) Suppose that the Cs atom at the comer A is missing. What is the net force now on the Cl atom due to seven remaining Cs atoms?
This is a Short Answer Type Questions as classified in NCERT Exemplar
Explanation- (i)The cesium atoms, are situated at the corners of a cube and Cl atom is situated at the centre of the cube. From the given figure, we can analyse that the chlorine atom is at equal distance from all the eight comers of cube where cesium atoms are placed. Thus, due to symmetry the electric field due to all Cs atoms, on Cl atom will cancel out. Hence net electric field at the centre of cube is zero.
(ii) we know, f=qE E= Kq/r2= Ke/r2
F= e (E)=e (ke/r2)=ke2/r2
Distance== 10-9m
F=8.99109 (1.610-16)2/ (0.34610-9)2=1.9210-9N
<p>This is a Short Answer Type Questions as classified in NCERT Exemplar</p><p><strong>Explanation- (i)</strong>The cesium atoms, are situated at the corners of a cube and Cl atom is situated at the centre of the cube. From the given figure, we can analyse that the chlorine atom is at equal distance from all the eight comers of cube where cesium atoms are placed. Thus, due to symmetry the electric field due to all Cs atoms, on Cl atom will cancel out. Hence net electric field at the centre of cube is zero.</p><p><strong> (ii) </strong>we know, f=qE E= Kq/r<sup>2</sup>= Ke/r<sup>2</sup></p><p>F= e (E)=e (ke/r<sup>2</sup>)=ke<sup>2</sup>/r<sup>2</sup></p><p>Distance=<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span>= <span contenteditable="false"> <math> <mroot> <mrow> <mrow> <msup> <mrow> <mrow> <mn>0.2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mn>0.2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mn>0.2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow></mrow> </mroot> <mo>*</mo> </math> </span>10<sup>-9</sup>m</p><p>F=8.99<span contenteditable="false"> <math> <mo>*</mo> </math> </span>10<sup>9</sup> (1.6<span contenteditable="false"> <math> <mo>*</mo> </math> </span>10<sup>-16</sup>)<sup>2</sup>/ (0.346<span contenteditable="false"> <math> <mo>*</mo> </math> </span>10<sup>-9</sup>)<sup>2</sup>=1.92<span contenteditable="false"> <math> <mo>*</mo> </math> </span>10<sup>-9</sup>N</p>
The following are the topics covered in this chapter: Electric Field and Field Lines, Gauss's Law, Electric Dipole, Conductors and Insulators, and Electric Flux.
Indeed, it is an easy chapter of Class 12 Physics. In this chapter, you will learn about the foundational concepts like Gauss's law and electric fields.
In the medical entrance test NEET, this chapter has a moderate weightage. You can expect around 2-3 questions of this chapter that contributes to the 3-5% of the total marks in the Physics section.
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