{"id":919,"date":"2024-07-27T05:57:05","date_gmt":"2024-07-27T05:57:05","guid":{"rendered":"https:\/\/www.meniit.com\/study-material\/?p=919"},"modified":"2024-10-04T05:33:19","modified_gmt":"2024-10-04T05:33:19","slug":"gausss-law","status":"publish","type":"post","link":"https:\/\/www.meniit.com\/study-material\/jee\/class-xith\/11th-physics\/gausss-law","title":{"rendered":"Gauss’s Law"},"content":{"rendered":"

Gauss law relates the flux through a closed surface (a surface that encloses some volume) with charges present inside the surface.\u201c
\nConsider a point charge q surrounded by a spherical surface of radius r<\/em> centered on the charge. The magnitude of the electric field everywhere<\/p>\n

on the surface on the sphere is E = k q\/r24<\/sup><\/p>\n

\"gaussian<\/p>\n

The electric field is perpendicular to the spherical surface at all points on the surface. The electric flux through the surface is therefore EA, where A = 4\u03c0r\u00b2 is the surface area of the sphere:<\/p>\n

\"formula\"<\/p>\n

\"note\"<\/p>\n

 <\/p>\n

\n

Example : <\/strong><\/h4>\n

Find out flux through the given Gaussian surface.
\n\"formula\"\"solution\"<\/p>\n<\/div>\n

\n

Example : <\/strong><\/h4>\n

If a point charge q is placed at the centre of a cube, then find out flux through any one face of cube.
\n\"solution\"<\/p>\n<\/div>\n

\n
\n

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