Answer
The electric potential at point P is $~~-1.78~V$
Work Step by Step
We can write a general equation for the net electric potential at a point:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
We can find the distance $r$ from the circumference of the circle to the point P:
$r = \sqrt{(8.20~cm)^2+(6.71~cm)^2}$
$r = 10.6~cm$
We can find the electric potential at point P:
$V_{net} = \frac{1}{4\pi~\epsilon_0}~(\frac{Q_1}{r}-\frac{6Q_1}{r})$
$V_{net} = \frac{1}{4\pi~\epsilon_0}~(-\frac{5Q_1}{r})$
$V_{net} = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~[-\frac{(5)(4.20\times 10^{-12}~C)}{0.106~m}]$
$V_{net} = -1.78~V$
The electric potential at point P is $~~-1.78~V$.
