Question

A charge of 6.65 mC is placed at each corner of a square 0.310 m on a side.

Part A : Determine the magnitude of the force on each charge.

Part B : Determine the direction of the force on a charge.

a)along the line between the charge and the center of the square toward the center

b)along the side of the square toward the other charge that lies on the side

c)along the side of the square outward of the other charge that lies on the side

d)along the line between the charge and the center of the square outward of the center

(please explain)

Answer 1

A)

q1, q2, q3,q4 –from the lower left corner – counterclockwise.

Origin of the coordinate system is at the corner where q1 is located.

F12=k•q1•q2/a² =

=9•10^9•6.65•10^-3•6.65•10^-3/0.31²=

= 4141545 N

F12x=0, F12y= - 4141545 N.

F13=k•q1•q3/ (a√2)² =

=9•10^9•6.65•10^-3•6.65•10^-3/0.31²•2=

= 2070773 N

F13x=F13y= - 2070773 •cos45= - 1464257 N.

F14=k•q1•q4/a² =

=9•10^9•6.65•10^-3•6.65•10^-3/0.31²=

= 4141545 N

F14x= - 4141545 N, F14y =0 N.

F1x= F12x+ F13x +F14x =0 - 1464257 - 4141545 = - 5605802 N,

F1y= F12y+ F13y +F14y = - 4141545 – 1464257 -0= - 5605802 N.

F1=sqrt(F1x² + F1y²) = 7927801 N.

B)

tanα = F1y/ F1x= 1

=> along the diagonal of the square, away from each charge.

The same force acts on each charge.

d)along the line between the charge and the center of the square outward of the center