Question

Two very small +3.00-μC charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the center of the meter stick.

Answer 1

Answer:

The electric potential at the center of the meter stick is 54 KV.

Explanation:

Electric potential (V) is given as:

i.e V = $\frac{kq}{r}$

Where: k is the Coulomb constant, q is the charge and r is the distance.

Given: q = 3.0 μC = 3.0 x $10^{-6}$ C, r = 0.5 m

So that,

V = $\frac{9*10^{9}*3.0*10^{-6} }{0.5}$

   = $\frac{2.7*10^{4} }{0.5}$

V = 54000

  = 54 000 volts

The electric potential at the center of the meter stick is 54 KV.