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
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,
1)
Given that, x = 1.10 m
y = 0.400 m
z = 0
Magnitude of x-component of electric field ,
Magnitude of y-component of electric field ,
∴
Magnitude of Electric field ,
2)
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