Question

Constants Part A Coulomb's law for the magnitude of the torce F between twco parsicles with charges Q and Q separated by a distance d is What is the net orce exerted by these two charges on a thied charge 9s-50.5 nC placed betweeng Your answer may be positive or negative, depending on the direction of the force. Express your answer numerieally in newtons to three significant flgures. View Available Hints) and -1.130? where 8854 x 10-* c/(N m2) s the permittivity of tree space Consider two point charges located on the x axis: one charge, 1125 DC, is located at -1.710 m; the second charge, 9s-30.0 nC is at the origin (Z-0.0000) Fore on qs--32.106 Submit incorrect; Try Again; 11 attempts remaining

Answer 1

Given : q₁ = -12.5 nC , r₁ = -1.710 m , q₂ = 38 nC , r₂=0 m ,

q₃ = 50.5 nC , r₃ = -1.130 m

Constant : K = 9×10⁹ N

Solution:

Part A Net force on charge q₃ due to q₁ and q₂

As we know that , the force between two charges q and q' is given by:

F = K|qq'|/r² , where r is the distance between two charges.

Distance between charge q₁ and q₃ is given by:

r = r₁-r₃ = -1.710-(-1.130) = 0.58 m

Force on charge q₃ due to q₁ is given by:

F₁₃ = K|q₁q₃|/r²

= (9×10⁹)(-12.5×10^-9)(50.5×10^-9)/(0.58)²

= 1.69 × 10^-5 N ( it is direcring toward negative x axis )

Distance between charge q₂ and q₃ is 1.130 m

Force on charge q₃ due to q₂ is given by:

F₂₃ = K(q₂q₃)/r²

=(9×10⁹)(38×10^-9)(50.5×10^-9)/(1.130)²

= 1.35×10^-5 N (it is directing toward negative x axis)

Net force acting on charge q₃ :

F_net = F₁₃+F₂₃

= (1.69×10^-5 N)+(1.35×10^-5 N)

= 3.04 × 10^-5 N

As it is directed toward negative x axis so , its sign will be negative.

Hence F_net = -3.04×10^-5 N