How does the electric field flux due to a point charge enclosed by Gaussian surface get
affected when its radios is double?
Answers
Answered by
2
Explanation:
ᴇʟᴇᴄᴛʀɪᴄ ғʟᴜx, ϕ=
ϵ
ᴏ
ǫ
ᴇɴᴄʟᴏsᴇᴅ
ᴀs ᴇʟᴇᴄᴛʀɪᴄ ᴅɪᴘᴏʟᴇ ᴄᴏɴsɪsᴛs ᴏғ ᴇǫᴜᴀʟ ᴘᴏsɪᴛɪᴠᴇ ᴀɴᴅ ɴᴇɢᴀᴛɪᴠᴇ ᴄʜᴀʀɢᴇs sᴇᴘᴀʀᴀᴛᴇᴅ ʙʏ sᴍᴀʟʟ ᴅɪsᴛᴀɴᴄᴇ. ᴛʜᴜs ɴᴇᴛ ᴄʜᴀʀɢᴇ ᴏɴ ᴀ ᴅɪᴘᴏʟᴇ ɪs ᴢᴇʀᴏ.
∴ ᴄʜᴀʀɢᴇ ᴇɴᴄʟᴏsᴇᴅ ʙʏ ᴛʜᴇ ᴄᴜʙᴇ, ǫ
ᴇɴᴄʟᴏsᴇᴅ
=
⟹ ϕ=
ϵ
ᴏ
=
Answered by
1
Answer:
That is, on increasing the radius of the gaussian surface, charge q remains unchanged. So, flux through the gaussian surface will not be affected when its radius is increased
Explanation:
hope it will be hlp full
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