Question

1. Consider three point charges at the corners of a triangle as shown in the figure, where q1=6.0x10-9 C , q3= 5.0x10-9 C, q2=-2.0x10-9 C, and the distance of separation are shown in the figure. Calculate the resultant force acting on q3. (Given: 60degree angle at q1 & q2, radius= 5 meters)

please explain. Thank you :)

Answer 1

So q3 doesn't generally go off into space, it just appears that it wouldbecause of q1, however truly there is an equivalent state among allcharges and it will be with in some limited separation of one another.

Presently to respond to your inquiry, you just need to utilize both sin and cos,because the outcome power is a blend of transgression and cos. Along these lines, toset this up you do every one of the powers in the x-bearing then all theforces in the y.

Fx = k (cos 60) ((q1)(q2)(52 ) +(q2)(q3)/(52 ) +(q1)(q3)(52 ))

Fy = k (sin 60) ((q1)(q2)(52 ) +(q2)(q3)/(52 ) +(q1)(q3)(52 ))

you have to connect every one of the qualities in to the conditions and youshould get:

Fx= k(cos( 60)) (1/25)((- 1.2x 10-17) + - 1 x10-17 +( 3x10-17))

Fx= k cos(60) (1/25)(8 x 10-18 )

=.27 x 10-8

Fy= k(sin( 60)) (1/25)((- 1.2x 10-17) + - 1 x10-17 +( 3x10-17))

Fy= k sin(60)(1/25) (8x - 18)

Fy= .87 x10-9

Presently to locate the resultant you have to utilize pythagoriam's thm

F= √(Fy2 +Fx2)

F= √(.87 x10-9 )2 +(.27 x10-8 )2)

F=√2.7 x 10-9

=5.23x 10-5