Three charges are fixed in place as shown. The squares in the grid have sides of

Question

Three charges are fixed in place as shown. The squares in the grid have sides of length s = 0.24 m. The magnitude of qis 3 ?Coulombs (1 ?Coulomb = 10-6 Coulombs), while the magnitude of Q is 4.5 ?Coulombs. What is the magnitude of the net force on q due to the other two charges?

F+q =

Three charges are fixed in place as shown. The squares in the grid have sides of length s = 0.24 m. The magnitude of q is 3? Coulombs (1 ?Coulomb = 10-6 Coulombs), while the magnitude of Q is 4.5 ?Coulombs. What is the magnitude of the net force on q due to the other two charges? F+q =

Explanation / Answer

force on q due to +Q = (kqQ/8s^2)*(1/sqrt(2) i + 1/sqrt(2) j)

where i and j are unit vectors along x-axis and y-axis

k = 9*10^9

force on q due to -Q = kqQ/4s^2 (-i)

So, net force on q , Fq = (kqQ/8s^2)*(1/sqrt(2) i + 1/sqrt(2) j) + kqQ/4s^2 (-i)

So, Fq = (kqQ/4s2)*((1/2*sqrt(2)-1) i + 1/(2*sqrt(2)) j)

So, Fq = (9*10^9*3*10^-6*4.5*10^-6/(4*0.24^2))*(-0.646 i + 0.354 j)

So, Fq = 0.527*(-0.646 i + 0.354 j)

So, magnitude of Fq = 0.527*sqrt(0.646^2+0.354^2) = 0.388 N <---------answer

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