Hi there!
Recall the equation for electric potential of a point charge:
[tex]V = \frac{kQ}{r}[/tex]
V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)
Q = Charge (C)
r = distance (m)
We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.
Upper right charge's potential:
[tex]V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V[/tex]
Lower left charge's potential:
[tex]V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V[/tex]
Add the two, and subtract from the total EP at the point:
[tex]3160 + 1123.75 - 2247.5 = 2036.25[/tex]
The remaining charge must have a potential of 2036.25 V, so:
[tex]2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}[/tex]
