Calculate the voltage at a point

[Colts]

Homework Statement

Calculate the electric potential at (a) point P in the first figure, and (b) point M in the second.
http://www.usi.edu/science/physics/pickett/206/5p3f1.jpg

Homework Equations

V=$\frac{q}{4\piεr}$
E=$\frac{q}{4\piεr^{2}}$

The Attempt at a Solution

I'm not sure if E is really necassaryt, but I wasn't sure how else to get the direction right. So I did the vector stuff with E and got
$\vec{E}$=<0,$\frac{-q}{2\pi\epsilon(.75)a^{2}}$>

Now I'm stuck and don't know how to get to voltage

 

[ap123]

You don't need the electric field - just use the definition of V.

 

[Colts]

Is it as easy as adding the two V's up?
Like
$V_{1}$+$V_{2}$=V
So the answer would be
V=$\frac{q}{2\pi\epsilon a}$

Does direction not matter?

 

[ap123]

Yes, you just add them up.
The potential is a scalar, not a vector, so it doesn't have a direction.

 

[Nickie]

.

To calculate the voltage at a point, you need to use the equation V = q/4πεr, where q is the charge at the point and r is the distance from the point to the source of the electric field. In this case, for point P, the distance r is 0.75a and the charge q is -3.5μC. Plugging in these values, we get:

V = (-3.5μC)/(4π(8.85x10^-12)(0.75a)) = -1.57x10^10 a^-1 V

For point M, the distance r is a and the charge q is 2.5μC. Plugging these values into the equation, we get:

V = (2.5μC)/(4π(8.85x10^-12)(a)) = 7.96x10^10 a^-1 V

Note that the negative sign for point P indicates that the voltage is negative, meaning the point is at a lower potential compared to infinity. The positive sign for point M indicates that the voltage is positive, meaning the point is at a higher potential compared to infinity.