Find the magnitude and sign of the point charge at the origin

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To find the magnitude and sign of the point charge at the origin, the electric field equation E = (KE * Q) / r^2 is used, where E is the electric field strength, KE is the Coulomb constant, and r is the distance from the charge. Given an electric field of 75.3 N/C at a distance of 82.8 m, the calculation results in a charge magnitude of approximately 5.743E-5 C. The direction of the electric field, which is radially inward, indicates that the charge is negative. Thus, the point charge at the origin is -5.743E-5 C.
bastige
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Homework Statement


The value of the Electric field at a distance of 82.8 m from a point charge is 75.3 N/C. Its direction is radially in toward the charge. The Coulomb constant is 8.98755 E9.
Find the magnitude and sign of the point charge at the origin. Answer in units C.


Homework Equations


I don't think I have all the necessary formulas. But this is what I do have
E=(KE*Q)/r^2 <---equation 1

F=[KE(Q1)(Q2)]/r^2 <--equation 2


The Attempt at a Solution



I am basically physics illiterate...so bare me some slack...
I used equation 1 to determine Q = 5.743E-5 .
How do I finish this problem? i feel like I'm missing a formula, or need some extra information.
 
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Coulomb's Law

\mid \underline{E} \mid=\frac{1}{4 \pi \epsilon_{0}}\frac{Q}{r^2}=\left[\frac{1}{C^2N^{-1}{m^-2}}\frac{C}{m^2}\right]

So that gives you the magnitude of Q, and the direction of the electric field lines tells you the sign of the charge.

It's a point charge so its that simple
 

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