Question

the electric field at the point x = 5.00cm and y = 0 points in the positive x direction with a magnitude of 10.0n/c. at point x = 10.0cm and y = 0, the electric field points in the positivex direction with a magnitude of 15.0n/c. assuming this electric field is produced by a single point charge, find its location.

a. 93cm
b. 45cm
c. 78cm
d. 32cm
e. 16cm

Explanation:

Step1: Recall electric - field formula

The electric - field due to a point charge $Q$ at a distance $r$ from the charge is given by $E = k\frac{Q}{r^{2}}$, where $k=\ 9\times10^{9}\ N\cdot m^{2}/C^{2}$. Let the point - charge be located at $x = a$. Then the electric field at $x_1 = 5.00\ cm$ is $E_1=k\frac{Q}{(a - 5)^{2}}$ and at $x_2 = 10.0\ cm$ is $E_2=k\frac{Q}{(a - 10)^{2}}$.

Step2: Set up the ratio of electric fields

We know that $\frac{E_1}{E_2}=\frac{(a - 10)^{2}}{(a - 5)^{2}}$. Given $E_1 = 10.0\ N/C$ and $E_2 = 15.0\ N/C$, so $\frac{10}{15}=\frac{(a - 10)^{2}}{(a - 5)^{2}}$. Taking the square - root of both sides, we get $\sqrt{\frac{10}{15}}=\frac{a - 10}{a - 5}$ (we consider the positive square - root since the electric field is in the positive $x$ - direction at both points, meaning the charge is to the right of both points). So, $\frac{\sqrt{10}}{\sqrt{15}}=\frac{a - 10}{a - 5}$. Cross - multiply: $\sqrt{10}(a - 5)=\sqrt{15}(a - 10)$.

Step3: Expand and solve for $a$

$\sqrt{10}a-5\sqrt{10}=\sqrt{15}a - 10\sqrt{15}$. Rearranging terms: $\sqrt{10}a-\sqrt{15}a=5\sqrt{10}-10\sqrt{15}$. Factoring out $a$: $a(\sqrt{10}-\sqrt{15})=5\sqrt{10}-10\sqrt{15}$. Then $a=\frac{5\sqrt{10}-10\sqrt{15}}{\sqrt{10}-\sqrt{15}}$. Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{10}+\sqrt{15}$:
\[

$$\begin{align*} a&=\frac{(5\sqrt{10}-10\sqrt{15})(\sqrt{10}+\sqrt{15})}{(\sqrt{10}-\sqrt{15})(\sqrt{10}+\sqrt{15})}\\ &=\frac{5\times10 + 5\sqrt{10\times15}-10\sqrt{15\times10}-10\times15}{10 - 15}\\ &=\frac{50+5\sqrt{150}-10\sqrt{150}-150}{- 5}\\ &=\frac{- 100 - 5\sqrt{150}}{-5}\\ & = 20+\sqrt{150}\approx20 + 12.25\\ &=32\ cm \end{align*}$$

\]

Answer:

D. $32\ cm$