Physics, asked by GINITHOMAS3292, 8 hours ago

In a certain region of space the electric potential is given by V=+Ax2y−Bxy2, where A = 5.00 V/m3 and B = 8.00 V/m3.
1) Calculate the magnitude of the electric field at the point in the region that has cordinates x = 1.10 m, y = 0.400 m, and z = 0.
2)Calculate the direction angle of the electric field at the point in the region that has cordinates x = 1.10 m, y = 0.400 m, and z = 0.( measured counterclockwise from the positive x axis in the xy plane)

Answers

Answered by malavikathilak123
11

Answer:

1)The magnitude of the electric field at the point in the region that has coordinates x = 1.10 m, y = 0.400 m, and z = 0, is 3.27 N/C

2)The angle of the electric field at the point in the region that has coordinates x = 1.10 m, y = 0.400 m, and z = 0, is 107.6 degrees

Explanation:

Given that,

     Electric potential,  V=Ax^2y-Bxy^2

                                     A=5.00 V/m^3\\B=8.00 V/m^3

1)  E=-\nabla V

       =E_xi+E_yj

   E_x=\frac{-\partial V}{\partial x}=-2Axy+By^2

   Ey=-\frac{\partial V}{\partial y}=-Ax^2+2Bxy

   Given that, x = 1.10 m

                       y = 0.400 m

                       z = 0

   Magnitude of x-component of electric field ,

                        \begin{aligned}E_x=\left[-2\times 5\times 1\cdot 1\times 0\cdot 400\right]+\left[8\times 0\cdot 400^2\right]\\=-3.12N/C\end{aligned}

   Magnitude of y-component of electric field ,

                       \begin{array}{l}E_y=\left[-5\times 1\cdot 10^2\right]+\left[2\times 8\times 1\cdot 10\times 0.40\right]\ =\ 0.99 V/C\end{array}

  ∴ E=-3\cdot 12i+0.99j

     Magnitude of Electric field ,

                            \begin{aligned}E=\sqrt{E_x^2+E_y^2}=\sqrt{\left(-3.12\right)^2+\left(0\cdot 99\right)^2}\\=3\cdot 27N/C\end{aligned}

2)      E=-3\cdot 12i+0.99j

         \theta =\tan ^{-1}\frac{E_x}{E_y}=\tan ^{-1}\frac{0\cdot 99}{3\cdot 12}=17\cdot 6

        Since x-component is negative and y-component is positive, the angle lies in second quadrant.

Therefore the angle of the electric field = 90 + 17.6 = 107.6 degrees

         

         

   

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