Question

A charge Q is located at a grid intersection in the x-y plane (see plot below), with Q=3.30 mu C The electric potential in the x-y plane due to a charge distribution near the origin is given by V(x, y) = beta x/(x^2+y^2)^3/2, with beta=5.347 times 10^5 Nm^3/C, x and y in m. Calculate the x-component of the electric force on Q. Part B The electric potential V(x, y) is calculated at the locations N and U using the equation above. Its values are V_N=8.956 times 10^3 Volts and V_U=8.167 times 10^3 Volts. Calculate the y-component of the force on Q.

Answer 1

A charge Q is located at a grid intersection in the x-y plane which is given below as :

4,4 4.2 3.8 5.6 5.8 6 6.26.4 x (n)

The electric potential in the x-y plane due to a charge distribution near the origin is given by,

V (x,y) = _$\beta$ x / (x² + y²)^3/2

where, _$\beta$ = 5.347 x 10⁵ Nm³/C

x = 0.4m & y = 0.4m

then, we get

V = (5.347 x 10⁵ Nm³/C) (0.4 m) / [(0.4 m)² + (0.4 m)²]^3/2

V = (2.138 x 10⁵ Nm⁴/C) / (0.32 m²)^3/2

V = (2.138 x 10⁵ Nm⁴/C) / (0.181 m³)

V = 1.18 x 10⁶ V

Part-A : The x-component of the electric force which will be given as :

F_x = q E

where, E = electric field = V / d

F_x = q (V / d) = (3.3 x 10^-6 C) (1.18 x 10⁶ V) / (0.565 m)                           (d = _$\sqrt{}$x² + y²)

F_x = 6.89 N

Part-B : The y-component of force on Q will be given as -

using a formula, we have

W = q $\Delta$V

F_y d = q (V_N - V_U)

F_y = (3.3 x 10^-6 C) [(8.956 x 10³ V) - (8.167 x 10³ V)] / (0.565 m)

F_y = 4.60 x 10³ N