Question

Point charges of 30.0 µC and 44.0 µC are placed 0.500 m apart.

Answer 1

To solve this problem we will apply the mathematical concepts given in Coulomb's laws. We will start from the definition of the potential difference, which is understood as the product between the Coulomb constant by the load on the distance between the two objects.

Later we will apply the concept of energy, which is similarly understood as the product between the charge squared by the Coulomb constant over the distance squared.

The voltage difference would be:

$V = \frac{kq}{r}$

Here,

k = Coulomb's constant

q = Charge

r = Distance

At the same time energy can be defined as,

$E = \frac{kQ_1Q_2}{r^2} = \frac{kQ^2}{r^2}$

PART A) For this part the potential difference at two objects is the same, then

$V_1 = V_2$

$\frac{k(30)}{x^2} = \frac{k(44)}{(0.5-x)^2}$

$\frac{(30)}{x^2} = \frac{(44)}{(0.5-x)^2}$

Solving for x,

$x = 0.2261m$ from $30\mu C$

Part b)  The magnitude (in N/C) and direction of the electric field halfway between them is

E at $x = 0.2261m$

$|E| = \frac{k(30*10^{-6})}{(0.2261)^2}-\frac{k(44*10^{-6})}{(0.2261)^2}$

$|E| = \frac{(9*10^{9})(30*10^{-6})}{(0.2261)^2}-\frac{(9*10^{9})(44*10^{-6})}{(0.2261)^2}$

$|E| = 2.4647*10^6 N/C$