Question

hi! i am having so much trouble understanding 2 problems from our module, and i would just like some guidance from anyone here! any help is appreciated! thank you in advance and sorry for the trouble.

first is:
A particle of charge 3x10^-9 C is located at point (0,0) in a certain coordinate system. Assuming all lengths are in meters, calculate the potential at the following points: A (3,3) and B (0,2). How much work is needed to take a particle of charge 2x10^-5 C from point A to point B?

and the second one is:
Equal charges of 3x10^-9 C are situated at the three corners of a square of side 5.20 m. Find the potential at the unoccupied corner.

Answer 1

the first one: Potential is V=kq/distance q is the charge, distance is frm the origin to pt A or B.
so the distance from 0,0 to 3,3 is sqrt (9+9) or 3 sqrt2 You know k, and q.
work needed from A to B is (Vb-Va)q

the second Same as the first, except you have different distances, you add them, potenial is not a vector
V= (V1 + V2 + V3) where the Vi is the potential at the unoccupied corner due to the charge at the first corner.

Answer 2

thank you so much! i think i can understand it better now :))

Answer 3

First the really easy part.
Once you have the potential (Voltage) at A and at B
THAT IS the work needed to bring a ONE Coulomb charge from A to B.
B is closer to the + charge at (0,0) so it will take positive work to move a plus charge from A to B with a plus moving charge in your hand.
Now A is how far from (0,0)?
sqrt (3^2+3*2) = 3 sqrt 2
Voltage (potential) at A = 9*10^9 (3*10^-9) / 3sqrt2
Voltage (potential) at B = 9*10^9 (3*10^-9) / 2
work done = charge * change in voltage = 2*110^-5 (Vb-Va)