Answer:
The electric potential at the center of the meter stick is 54 KV.
Step-by-step explanation:
Electric potential (V) is given as:
i.e V =
![(kq)/(r)](https://img.qammunity.org/2022/formulas/physics/college/me9mgcsajqhx60ewren0x70178e5p0radf.png)
Where: k is the Coulomb constant, q is the charge and r is the distance.
Given: q = 3.0 μC = 3.0 x
C, r = 0.5 m
So that,
V =
![(9*10^(9)*3.0*10^(-6) )/(0.5)](https://img.qammunity.org/2022/formulas/physics/college/q8cg6cetjkgf36f423mtawb0awib6q59tc.png)
=
![(2.7*10^(4) )/(0.5)](https://img.qammunity.org/2022/formulas/physics/college/v1cjh1u8vbsedme4qjib8q8uszztbqcxym.png)
V = 54000
= 54 000 volts
The electric potential at the center of the meter stick is 54 KV.