Point charge and electric potential

[uniidiot]

Homework Statement

three point charges, each of 10^-9 are located at 3 corners of a square of side 20cm. Calculate the electric potential at the forth corner.

Homework Equations

V = Q / (4(pi)(epsilon 0) r)

The Attempt at a Solution

Potential because of corner A = 10^-9 / (4 (pi) (epsilon 0) 0.2)

= 44.96

Potential because of corner B = 10^-9 / (4 (pi) (epsilon 0) 0.2)

= 44.96

Potential because of corner C = 10^-9 / (4 (pi) (epsilon 0) 0.284)

= 31.66

A + B + C = 121.58


Is this correct, finding the potential from each point charge individualy?

Also there is another question very similar to this where there are 4 point charges on the corners of a square, and we need to find the potential at the centre, would i do this the same as above, (providing what i did above is correct.

 

[nrqed]

Homework Statement

three point charges, each of 10^-9 are located at 3 corners of a square of side 20cm. Calculate the electric potential at the forth corner.

Homework Equations

V = Q / (4(pi)(epsilon 0) r)

The Attempt at a Solution

Potential because of corner A = 10^-9 / (4 (pi) (epsilon 0) 0.2)

= 44.96

Potential because of corner B = 10^-9 / (4 (pi) (epsilon 0) 0.2)

= 44.96

Potential because of corner C = 10^-9 / (4 (pi) (epsilon 0) 0.284)

= 31.66

A + B + C = 121.58


Is this correct, finding the potential from each point charge individualy?

Also there is another question very similar to this where there are 4 point charges on the corners of a square, and we need to find the potential at the centre, would i do this the same as above, (providing what i did above is correct.

Yes, it's correct (you should put the units on your answers, though!). I get a slightly different value for the distance for the third chare (0.2828 m which is approximately 0.283) . You are doing it the right way: finding the potential due to each charge and then adding all the results.

 

[uniidiot]

thank you.

I get a slightly different value for the distance for the third chare (0.2828 m which is approximately 0.283)

i think it was a typing error lol.