Question

In the figure below, each charged particle is located at one of the four vertices of a square with side length In the figure, A = 5, B=2, and C=6, and q > 0. 

a) What is the expression for the magnitude of the electric field in the upper right corner of the square (at the location of q)? 

Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric field at this location. 

(b) Determine the expression for the total electric force exerted on the charge q.

Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric force on q.

Answer 1

The Electric field on q on account of charge 5q = ke (5q) /a^2 ---- along +x-direction

The Electric field on q on account of charge 6q = ke (6q) /a^2 -------- along +y-direction

The Electric field on q on account of charge 2q = ke (2q) / 2a^2 ........ along diagonal joining Bq to q and away from q and Bq

The third field above is at 45 deg with respect to X-axis

So component of this field along +x-direction = [(ke)(2q)/a^2] cos 45 = (ke) ( sqrt 2)(q )/ ( a^2)

component along +Y-direction = (ke)( q ) (sqrt 2) / ( a^2 )

Totla field along +X-axis = [(ke)(5q) /a^2} + (ke)(sqrt2)q / a^2 ]

= [(ke)(q)/a^2 ] [ 5 + (sqrt 2) ] = 6.41[ (ke) (q) /a^2]

Total field along + Y-direction = Ke q /a^2 [ 6+ sqrt2] = 7.41[(ke) q/a^2]

Resultant Field = [(keq/a^2]Sqrt ( 6.41^2 + 7.41^2 ) = 9.8 Ke q / a^2

= 49.1deg with respect +x-axis

(b) Total force on q is = q E(total) = 9.8 (ke) (q) (q) /a^2 = 9.8 (ke) (q^2) /a^2 and its direction is the same as that of the net E field, which is 49.1 deg with respect to +X-axis