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प्रश्न
The potential at a point x (measured in µm) due to some charges situated on the X-axis is given by v(x) = `20/((x^2 - 4)` V. The electric field E at x = 4 µm is given by ______.
पर्याय
`5/2` V/µm and in the -ve x-direction
`5/2` V/µm and in the +ve x-direction
`10/9` V/µm and in the -ve x-direction
`10/9` V/µm and in the +ve x-direction
MCQ
रिकाम्या जागा भरा
उत्तर
The potential at a point x (measured in µm) due to some charges situated on the X-axis is given by v(x) = `20/((x^2 - 4)` V. The electric field E at x = 4 µm is given by `underlinebb(10/9 "V""/"µ"m" "and in the +ve x-direction")`.
Explanation:
E = `-(∂V)/(∂x)hati - (∂V)/(∂y)hatj - (∂V)/(∂z)hatk`
⇒ `E_x = -(∂V)/(∂x) = -d/dx[20/(x^2 - 4)] = (40x)/((x^2 - 4)^2)`
⇒ Ex at x = 4 µm = `10/9` V/µm and is along +ve x-direction.
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