Question

The electric potential for various points in x-y plane is given by V = 1.0 x²   − 2.0 y², where V is in volts and x, y are in metres. The angle that the electric field at point (2.0 m, 1.0 m) makes with the positive x-axis is ______.

पर्याय

• 45°

• 90°

• 135°

• 315°

Answer 1

The electric potential for various points in x-y plane is given by V = 1.0 x²   − 2.0 y², where V is in volts and x, y are in metres. The angle that the electric field at point (2.0 m, 1.0 m) makes with the positive x-axis is 135°.

Explanation:

The given potential V (interpreted from the provided text as V = 1.0 x²   − 2.0 y²) depends on the coordinates x and y.

The electric field `vec E` is the negative gradient of the electric potential, defined by the components:

E_x = `-(delta V)/(delta x)`

E_y = `-(delta V)/(delta y)`

Applying these to the function:

E_x = `-delta/(delta x)(1.0 x^2 - 2.0 y^2)`

= −2x

E_y = `-delta/(delta y)(1.0 x^2 - 2.0 y^2)`

= 4y

Substitute the co-ordinates (x, y) = (2.0, 1.0) into the component equation:

E_x = −2(2.0)

= −4.0 V/m

E_y = 4(1.0)

= 4.0 V/m

The angle θ with the positive x-axis is found using the tangent ratio:

tan(θ) = `E_y/E_x`

= `4.0/-4.0`

= −1

Since E_x is negative and E_y is positive, the electric field vector lies in the second quadrant. The angle θ is:

θ = tan⁻¹(−1)

= 135°