Question

What is the force on the charge q at the lower-right-hand corner of the square shown here? (Use the following as necessary for the magnitude: q, a and Eo Assume that the +x-axis is to the right and the ty-axis is up along the page. Enter the direction in degrees counterclockwise from the +x-axis.) magnitude F= direction 45 counterclockwise from the +x-axis

Answer 1

In triangle ABC

using pythagorean theorem

AC = sqrt(AB² + BC²) = sqrt((a)² + (a)²) = sqrt(2) a

$\theta$ = tan^-1(BC/AB) = tan^-1(a/a) = 45

F = force by charge at B and D on charge at C = k q²/a²

F' = force by charge at A on charge at C = k q²/(sqrt(2) a)² = (0.5)  k q²/a²

net force along the X-direction is given as

F_x = F + F' Cos$\theta$ = (k q²/a² ) + (0.5) Cos45 k q²/a² = (1.4) k q²/a²

net force along the Y-direction is given as

F_y = - F - F' Sin$\theta$ = -(k q²/a² ) - (0.5) Sin45 k q²/a² = - (1.4) k q²/a²

Net force is given as

F = sqrt(F_x² + F_y²) = sqrt(2) (1.4) k q²/a² = (1.98) k q²/a²

direction : $\phi$ = 360 - tan^-1(F_y/F_x) = 360 - tan^-1(F_y/F_x) = 360 - tan^-1(1) = 315