Question

The charges q1 =1.5mc q2=0.2mc q3=-0.5are placed at 3 different point ABC as shown in fig .if r1=1.2,r2=0.6, calculate the magnitude and the resultant force on q2​

Answer 1

force exerted by $q_1$ on $q_2$, F = $\frac{kq_1q_2}{r_1^2}$

= 9 × 10^9 × 1.5 × 10^-3 × 0.2 × 10^-3/(1.2)²

= 2.7 × 10³/1.44

= 1.875 × 10³ N

= 1875 N

force exerted by $q_3$ on $q_2$, F' = $\frac{kq_3q_2}{r_2^2}$

= 9 × 10^9 × 0.5 × 10^-3 × 0.2 × 10^-3/(0.6)²

= 0.9 × 10³/(0.36)

= 2.5 × 10³ N

= 2500N

now, resultant force on $q_2$ = √{F² + F'²}

= √{1875² + 2500²} N

= 3125 N

direction of resultant force :

in vector form , force exerted by $q_1$ on $q_2$ ,F= 1875 î N

in vector form, force exerted by $q_3$ on $q_2$ , F'= -2500 j N

so, direction of resultant force with x - axis = tan^{-1}{2500/1875} = tan^{-1}(4/3) with x-axis

Answer 2

Explanation:

please check out the above attachment

..TQ