Question

A charge particle q, is at position (2. - 1, 3). The electrostatic force on another charged particle q, at
(0, 0, 0) is :​

Answer 1

Answer:

kqq/14

Explanation:

the difference btw q is 14^1/2

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,

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

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

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