Why is the change in electric potential leading to a negative work value?

[Ry122]

http://users.on.net/~rohanlal/elec.jpg

for question b, wouldn't there be no change in electric potential since
electric potential for a particular point is determined by the sum of the electric potentials for
each particular charge about that point, and if a charge is moved to point p
and the eq for electric potential is kq/r then that charge would be 0 since r would be 0.

 

[LowlyPion]

for question b, wouldn't there be no change in electric potential...

In b) they are asking for the change in potential energy. This is not the same as asking for the electric potential - i.e. the voltage.

The change in potential energy is going to be expressed as the work required to move the charge.

For b) then what you want to explore is work given by ...

W = qΔV

The voltage at ∞ is 0, so ... ΔV = simply the voltage at P.

 

[Ry122]

According to my lecture notes electric potential is the same as potential energy.

the answer to this problem requires the initial electric potential of the system which is found by

http://users.on.net/~rohanlal/pot.jpg

to be subtracted from 3.0x10^-6
which gives a final answer of - 1.89 x 10-2

but i don't understand why this is the case

 

[diazona]

According to my lecture notes electric potential is the same as potential energy.

Your lecture notes are wrong, then...

Mathematically, electric potential is
$$V = k\frac{q}{r}$$
whereas electrical potential energy is
$$U_E = k\frac{q_1 q_2}{r}$$

 

[LowlyPion]

According to my lecture notes electric potential is the same as potential energy.

the answer to this problem requires the initial electric potential of the system which is found by

to be subtracted from 3.0x10^-6
which gives a final answer of - 1.89 x 10-2

but i don't understand why this is the case

It's not subtracted.

The Work is the charge times the change in voltage potential.

What they want is Work.

And if you multiply the charge by the Voltage, then you get the answer indicated.