1.Two point charges are fixed on the y axis: a negative point charge q1 = -27 µC
ID: 1951578 • Letter: 1
Question
1.Two point charges are fixed on the y axis: a negative point charge q1 = -27 µC at y1 = +0.24 m and a positive point charge q2 at y2 = +0.32 m. A third point charge q = +8.2 µC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 26 N and points in the +y direction. Determine the magnitude of q2.
2.At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 271 N/C. At a distance r2 from the charge, the field has a magnitude of133 N/C. Find the ratio r2/r1.
Explanation / Answer
Charges q 1 = -27 x 10 -6 C y 1 = 0.24 m q 2 = ? y 2 = 0.32 m q = 8.2 x 10 -6 C Force on q due to q 1 is F = K qq 1 / y 12 Where K = Coulomb's constant = 8.99 x 10 9 Nm 2 / C 2 Substitute values we get F = 34.55 N Force on q due to q 2 is F ' = K qq 2 / y 22 Where K = Coulomb's constant = 8.99 x 10 9 Nm 2 / C 2 Substitute values we get F ' = 719.902 x 10 3 q 2 Given F + F ' = 26 N F ' = 26 - F = 26-34.55 719.902 x 10 3 q 2 = (26-34.55) q 2 = -11.87 x 10 -6 C (b).we know E = Kq / r 2 From this E 1 / E 2= ( r 2/ r 1 ) 2 271 / 133 = ( r 2/ r 1 ) 2 ( r 2/ r 1 )= [271/133] = 1.427 Force on q due to q 2 is F ' = K qq 2 / y 22 Where K = Coulomb's constant = 8.99 x 10 9 Nm 2 / C 2 Substitute values we get F ' = 719.902 x 10 3 q 2 Given F + F ' = 26 N F ' = 26 - F = 26-34.55 719.902 x 10 3 q 2 = (26-34.55) q 2 = -11.87 x 10 -6 C (b).we know E = Kq / r 2 From this E 1 / E 2= ( r 2/ r 1 ) 2 271 / 133 = ( r 2/ r 1 ) 2 ( r 2/ r 1 )= [271/133] = 1.427Related Questions
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