The electric potential between two metallic plates in a diode is given by V(x) =

Question

The electric potential between two metallic plates in a diode is given by V(x) = Ax^4/3 where x is the distance in meters measured from one plate. the distance between the plates is 2.0cm, and the potential at the other plate is 1000V. Determine the constant A and the electric field at the point x = 1.0 cm On the x-y plane where coordinates are measured in meters, the electric potential at the point (x, y) is given by V(x, y) = 8x^2y-3y^3 (V) Find the x any y components of the electric field at the point (2, -1). On the x-y plane, two point charges q are located at the points (a.0) and (-a, 0). Write an expression for the electric potential at the point (x, y). Use it to find the x and y components of the electric field at the point (0, y) on the y-axis. Hence determine the y coordinate of the point on the y = axis where the electric field is maximum.

Explanation / Answer

1000 = A (0.02) ^ 4/3

A ( constant) = 1000/ 0.0055 = 181818.18 = 1.82 x 10^5 apprx

E =- dv/dx

E = 1.84 x 10^5 ( 4/3) x^ 1/3

E( 1cm) = 245333.33 ( 0.01) ^1/3 = -53673.08 N/C apprx

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