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प्रश्न
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 ______.
विकल्प
45°
90°
135°
315°
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उत्तर
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:
Ex = `-(delta V)/(delta x)`
Ey = `-(delta V)/(delta y)`
Applying these to the function:
Ex = `-delta/(delta x)(1.0 x^2 - 2.0 y^2)`
= −2x
Ey = `-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:
Ex = −2(2.0)
= −4.0 V/m
Ey = 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 Ex is negative and Ey is positive, the electric field vector lies in the second quadrant. The angle θ is:
θ = tan−1(−1)
= 135°
