Two charges q1 = ?3.20 nC and q2 = +8.31 nC are at a distance of...

Two charges q₁ = ?3.20 nC and q₂ = +8.31 nC are at a distance of 1.62

Expert Solution

a) distance travelled by charge q2 from 1.62 micro metre to 0.340 micro metre is = 1.62-0.340 = 1.28 *10^{-6} m

the force on q2 at 0.340 micro metre is, F = k*|q1*q2|/r^2

= 9*10^9 * 3.2*10^{-9}*8.31*10^{-9}/(0.34*10^{-6})^2

= 2.07*10^{6} N

from kinematic equation,

v^2 - u^2 = 2*a*S

here u = 0 , S = 1.28*10^{-6} m then

a = v^2/(2*S) = v^2/( 2*1.28*10^{-6} )

Since F= m*a = m*v^2/( 2*1.28*10^{-6} )

(1/2)*m*v^2 = F*1.28*10^{-6} = 2.6496 J

the kinetic energy is 2.6496 J

b) if m = 7.85*10^{-6} Kg then

acceleration,a = F/m = 0.2636*10^{12} m/s^2

then speed, v = sqrt(2*a*S) = sqrt(0.674*10^{6}) = 0.82*10^3 m/s

genius_generous answered 1 month ago

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