Electric Intensity Due To An Infinite Sheet Of Charge. Assume we have a plane sheet of infinite extent on which the positive charges are uniformly distributed.<\/p>\n\n\n\n
Now imagine a cylinder passing through this sheet.<\/p>\n\n\n\n
The charge enclosed by the surface is A. <\/p>\n\n\n\n
The first face of the infinite sheet<\/p>\n\n\n\n
1=E.Acos<\/p>\n\n\n\n
cos=1<\/p>\n\n\n\n
1=E.A<\/p>\n\n\n\n
Second face of the infinite sheet<\/p>\n\n\n\n
2=E.Acos<\/p>\n\n\n\n
cos=1<\/p>\n\n\n\n
2=E.A<\/p>\n\n\n\n
Third face of the infinite sheet<\/p>\n\n\n\n
3=E.Acos<\/p>\n\n\n\n
cos=0<\/p>\n\n\n\n
3=0<\/p>\n\n\n\n
The total net flux will be determined by<\/p>\n\n\n\n
=1+2+3<\/p>\n\n\n\n
=E.A+E.A+0<\/p>\n\n\n\n
=2E.A<\/p>\n\n\n\n
According to the Gauss\u2019s law<\/p>\n\n\n\n
=1Qt<\/p>\n\n\n\n
Since is the charge density, Q can be considered as Q=.A<\/p>\n\n\n\n
1A<\/p>\n\n\n\n
2E.A=1A<\/p>\n\n\n\n
A and A will be canceled out.<\/p>\n\n\n\n
This leads us to <\/p>\n\n\n\n
2E=1<\/p>\n\n\n\n
2E=<\/p>\n\n\n\n
E=2<\/p>\n","protected":false},"excerpt":{"rendered":"
Electric Intensity Due To An Infinite Sheet Of Charge. Assume we have a plane sheet of infinite extent on which the positive charges are uniformly distributed. Now imagine a cylinder passing through this sheet. The charge enclosed by the surface is A. The first face of the infinite sheet 1=E.Acos cos=1 1=E.A Second face of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47],"tags":[],"class_list":["post-25555","post","type-post","status-publish","format-standard","hentry","category-titan"],"yoast_head":"\n