[amp_mcq option1=”8.854 x 10-9 F\/m” option2=”8.854 x 10-10 F\/m” option3=”8.854 x 10-11 F\/m” option4=”8.854 x 10-12 F\/m” correct=”option1″]<\/p>\n
The correct answer is A. 8.854 x 10-9 F\/m.<\/p>\n
The absolute permittivity of free space, also known as the permittivity of vacuum, is a physical constant that is a measure of how much electric field is required to create a unit electric displacement in a vacuum. It is denoted by the Greek letter \u00ce\u00b50 and has the value 8.854 \u00c3\u0097 10\u00e2\u0088\u009212 farad per meter (F\/m).<\/p>\n
The permittivity of free space is a fundamental constant of nature and is not dependent on the properties of any particular material. It is used in many equations in electromagnetism, including Gauss’s law for electricity and Faraday’s law of induction.<\/p>\n
The permittivity of free space can be calculated from the speed of light in vacuum, c, and the electric constant, \u00ce\u00b50, as follows:<\/p>\n
\u00ce\u00b50 = 1\/(c2)<\/p>\n
where:<\/p>\n
The permittivity of free space is a very small number, which means that it takes a very large electric field to create a unit electric displacement in a vacuum. This is because the vacuum is a very good insulator.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”8.854 x 10-9 F\/m” option2=”8.854 x 10-10 F\/m” option3=”8.854 x 10-11 F\/m” option4=”8.854 x 10-12 F\/m” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[971],"tags":[],"class_list":["post-47873","post","type-post","status-publish","format-standard","hentry","category-electrostatics","no-featured-image-padding"],"yoast_head":"\n