Two point charges +8q and -2q are located at x=0 and x=L respectively. The point on x axis at which net electric field is zero due to these charges is-
(i) 8L
(ii) 4L
(iii) 2 L
(iv) L
(iii) 2 L
Explanation:
Superposition of Electric field = The resultant electric field at any point is equal to the vector sum of all the electric fields.
so,
`\vec{"E"} = \vec{"E"}_{1}+\vec{"E"}_{2}+\vec{"E"}_{3}+...\vec{"E"}_{n}`
`\frac{1}{4\pi\epsilon_{0}}\frac{"8q"}{(L+d)^{b}}-\frac{1}{4\pi\epsilon_{0}}\frac{2"q"}{"d"^{2}}`
`("L+d"^{2})` = `"4d"^{2}`
d = L
∴Distance from origin = 2L
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