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

[bastige]

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.

 

[neu]

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

 

[Moetasim]

As a scientist, it is important to have all the necessary information before attempting to solve a problem. In this case, you are correct in thinking that you are missing some information. In order to calculate the magnitude and sign of the point charge at the origin, we need to know the value of the electric field at a specific distance from the charge, not just the distance and direction.

In order to solve this problem, we need to use the equation for the electric field, E = (kQ)/r^2, where k is the Coulomb constant, Q is the charge of the point charge, and r is the distance from the charge. Using this equation, we can rearrange it to solve for Q, which will give us the magnitude and sign of the point charge.

Q = Er^2/k

Plugging in the given values, we get:

Q = (75.3 N/C)(82.8 m)^2/(8.98755 x 10^9 N*m^2/C^2)

Q = 5.743 x 10^-5 C

So, the magnitude of the point charge at the origin is 5.743 x 10^-5 C, which is the same value you calculated. However, in order to determine the sign of the charge, we need to know the direction of the electric field. Since the electric field is radially in towards the charge, we can determine that the charge must be negative, as the direction of the electric field is always away from positive charges and towards negative charges.

Therefore, the magnitude and sign of the point charge at the origin is 5.743 x 10^-5 C and negative, respectively. It is important to always double check your units and make sure they match up in order to ensure a correct solution.