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(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. |
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Answer» SOLUTION :(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` . |
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