(a) Calculate the electric potential at points P and Q as shown in the figure below.
(b ) Suppose the charge `+9muC` is replaced by `-9muC` find the electrostatic potentials at points P and Q

( c) Calculate the work done to bring a test charge `+2muC` from infinity to the point P. Assume the charge `+9muC` is held fixed at origin and `+2muC` is brought from infinity to P.

Explanation

(a) Electric potential at point P is given by `V_(p)=(1)/(4piepsilon_(0))(q)/(r_(p))=(9xx10^(9)xx10^(-6))/(10) =8.1xx10^(3)V`
Electric potential at point Q is given by `V_(Q)=(1)/(4piepsilon)(q)/(r_(Q))=(9xx10^(-6))/(16)=5.06xx10^(3)V`
Note that the electric potential at point Q is less than the electric potential at point P. If we put a positive charge at P it moves from P to Q . However if we place a negative charge at P it will move towards the charge `+9muC` . The potential difference between the points P and Q is given by
`DeltaV=V_(P)-V_(Q)=+3.04xx10^(3)V`
(b) Suppose we replace the charge `+9muC"by"-9muC` then the corresponding potentials at the points P and Q are `V_(p)=-8.1xx10^(3)V, V_(Q)=-5.06xx10^(3)V`
Note that in this case electric potential at the point Q is higher than at point P. The potential difference or voltage between the points P and Q is given by
`DeltaV=V_(p)-V_(Q)=-3.04xx10^(3)V`
(c) The electric potentail V at a point P due to some charge is defined as the work done by an external force to bring a unit positive charge from infinity to P. So to bring the q amount of charge from infinity to the point P work done is given as follows.
W=qV
`W_(Q)=2xx10^(-6)xx5.06xx10^(3)J=10.12xx10^(-3)J` .