Question

The electric potential at the dot in the figure is 3160 V. What is charge q?

Answer 1

Hi there!

Recall the equation for electric potential of a point charge:

`V = (kQ)/(r)`

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:

`V = ((8.99*10^9)(-5 * 10^(-9)))/(0.04) = -1123.75 V`

Lower left charge's potential:

`V = ((8.99*10^9)(5*10^(-9)))/(0.02) = 2247.5 V`

Add the two, and subtract from the total EP at the point:

`3160 + 1123.75 - 2247.5 = 2036.25`

The remaining charge must have a potential of 2036.25 V, so:

`2036.25 = ((8.99*10^9)(Q))/(√(0.02^2 + 0.04^2))\\\\2036.25 = ((8.99*10^9)Q)/(0.0447) \\\\Q = 0.000000010127 = \boxed{10.13nC}`