Question

distance between the two point charges +9e and +e is 16 centimeter.where we can keep a third charge q between them so that they will be in equilibrium?

Answer 1

Answer on Question#68705 – Physics – Electromagnetism

Distance between the two point charges $+9\mathrm{e}$ and $+\mathrm{e}$ is 16 centimeter, where we can keep a third charge $q$ between them so that they will be in equilibrium?

**Solution.** Let's draw a sketch of the placement of charges

Let $x$ – distance between charges $e$ and $q$. Hence distance between charges $9e$ and $q$. Since the charges are in equilibrium, the resultant force acting on each of the charges is zero. Let us consider the forces acting on the charge $q$. The interaction of point charges is described by the Coulomb law

$$F = \frac {k \cdot q _ {1} \cdot q _ {2}}{r ^ {2}}$$

where $r$ – distance between charges $q_{1}$ and $q_{2}$, $k = 8.99 \cdot 10^{9} \frac{Nm^{2}}{C^{2}}$.

Therefore force between charges $e$ and $q$ equal to

$$F _ {1} = \frac {k \cdot e \cdot q}{x ^ {2}}.$$

Therefore force between charges $e$ and $q$ equal to

$$F _ {2} = \frac {k \cdot 9 e \cdot q}{(1 6 - x) ^ {2}}.$$

As result $F_{1} = F_{2}\rightarrow \frac{k\cdot e\cdot q}{x^{2}} = \frac{k\cdot 9e\cdot q}{(16 - x)^{2}}\rightarrow \frac{1}{x^{2}} = \frac{9}{(16 - x)^{2}}\rightarrow \frac{16 - x}{x} = 3\rightarrow 16 - x = 3x.$

$4x = 16\rightarrow x = 4$ cm.

**Answer.** For the charge to stay in equilibrium, it is necessary to place a charge of $q$ at a distance of 4 cm from the charge $e$.

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