Electric Field. Having trouble with this. I keep getting -2.52 for 1) which is incorrect. Please show steps?

A charge of -4.2 µC is located at the origin; a charge of 5.6 µC is located at x = 0.2 m, y = 0; a third charge Q is located at x = 0.32 m, y = 0. The force on the 5.6 µC charge is 470.4 N, directed in the positive x direction.

1) Determine the charge Q.

2) With this configuration of three charges, where, along the x direction, is the electric field zero?

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  • 8 years ago
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    Call the origin point a, the point with the 5.6 µC charge point b, and the point with the unknown charge point c.

    1) From the force on the 5.6 µC,

    E = F/q tells us that the field at b is 84*10^6 N/C pointing toward +x.

    Call the field calculated above Enet. That's the total field at point b.

    Let Ea be the field at point b due to the charge at a and let Ec be the field at b due to charge Q, at c.

    So Ea + Ec = Enet = 84*10^6 N/C

    The field at b due to the -4.2 µC is

    Ea = k*q/r^2 = 9*10^9*(-4.2*10^-6) / (0.2^2) = -945*10^3 N/C

    -945*10^3 N/C + Ec = 84*10^6 N/C

    so

    Ec = 84.945*10^6 N/C

    Notice that the direction is toward point c. So charge Q is putting out a negative field, so for this next part we need the field to be -84.945*10^6 N/C.

    Ec = -84.945*10^6 N/C = k*q/r^2 = 9*10^9*Q/(0.12)^2

    Q = -84.945*10^6*(0.12)^2 / 9*10^9 = -0.1359*10^-3 = -135.9*10^-6 = -135.9 µC

    2) Enet = 0 at some point d.

    0 = k*qa/ra^2 + k*qb/rb^2 + k*Q/rc^2 where for each term you enter the charge at the 3 points and an expression for the distance from d to points a,b,c respectively. Don't have time for more.

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