Question

A charged particle q1 is at position (2,-1,3). The electrostatic force on another charged particle q2 at (0,0,0) is

Answer 1

Given, charged particle q₁ position (2, -1, 3) and q₂ position (0,0,0)
Let position vector of q₁ = OP = 2i - j + 3k
Position vector of q₂ = OQ = 0i + 0j + 0k
Now, vector PQ , {separation between charges } = OP - OQ
= 2i - j + 3k
Magnitude of PQ = √{2² + 1² + 3² } = √14 unit

Now, electrostatic force in vector form,
$F=\frac{KQ_1Q_2}{|r|^3}\bold{r}$
Here r is the separation between charges .
F = Kq₁q₂/|√14|³ (2i - j + 3k )

Magnitude of force = Kq₁q₂/14 N

Answer 2

Answer:

$\vec{F} = \dfrac{q_1q_2(-2\hat{i}+\hat{j}-3\hat{k})}{56\sqrt{14}\pi\epsilon_0}$

Explanation:

Given.

A charged particle q1 is at position (2,-1,3).

To find: The electrostatic force on another charged particle q2 at (0,0,0) is

Solution:

Position vector along q1 = 2i - j + 3k

Position vector along q2 = 0i + 0j + 0k

Now, Electrostatic force can be given as:

$\displaystyle\vec{F} = \dfrac{kq_1q_2}{r^3}(\vec{r})$

where, k = $\dfrac{1}{4\pi\epsilon_0}$

Substituting the values,

$\vec{F} = \dfrac{1}{4\pi\epsilon_0}\left(\dfrac{q_1q_2\left[(0-2)\hat{i}+\{0-(-1)\}\hat{j} +(0-3)\hat{k}\right]}{\left[\sqrt{(0-2)^2+(0+1)^2 +(0-3)^2}\right]}\right)$

$\vec{F} = \dfrac{q_1q_2}{4\pi\epsilon_0}.\dfrac{(-2\hat{i}+\hat{j}-3\hat{k})}{(\sqrt{4+1+9})^3}$

$\boxed{\vec{F} = \dfrac{q_1q_2(-2\hat{i}+\hat{j}-3\hat{k})}{56\sqrt{14}\pi\epsilon_0}}$