Question

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 ______.

Options

• `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

Answer 1

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)`

⇒ E_x at x = 4 µm = `10/9` V/µm and is along +ve x-direction.