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

Answer:

$\\\tt 6.42 \times 10^8 \times q_1 q_2 \ N$

Explanation:

Using the coordinates the distance between the particles,

$\\\tt \sqrt{(2-0)^2 + (-1+0)^2 + (3-0)^2 } \\\\\tt = \sqrt{ 4 +1 +9 } \\\\\tt = \sqrt{14} \ units$

Fore between them,

$\\\tt F = \dfrac{1}{4 \pi \epsilon_{\circ} } \dfrac{ q_1 q_2} {r^2 } \\\tt F= 9 \times 10^9 \times \dfrac{ q_1 q_2}{(\sqrt{14})^2 } N \\\tt = [9 \times 10^9 \times \dfrac{ q_1 q_2}{14 }] N \\\\\tt = 6.42 \times 10^8 \times q_1 q_2 N$

Answer 2

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$\maltese \: { \underline{ \underline{ \bf{Answer :-}}}}$

Given :

• A charged particle q1 is at position (2,-1,3)
• Another charged particle q2 at (0,0,0)

To Find:

• Electrostatic force

Explanation :

$\underline{ \tt \: Position \: \: vector \: \: \vec {d}}$

$\bf \: \vec{d} \: = (2 - 0) \hat{i} - (1 - 0) \hat{j} + (3 - 0) \hat{k} = 2 \hat{i} - \hat{j} \: + 3 \hat{k}$

$\bf \: {d}^{3} = { ((2 \hat{i} - \hat{j} + 3\hat{k}) \: . \: (2 \hat{i} - \hat{j} + 3\hat{k})) }^{ \frac{3}{2} }$

$\pink{ \bf{ {d}^{3} = 14 \sqrt{14}}}$

Electrostatic Force:-

$\bf \: F = \: \frac{1}{4\pi \epsilon_{ \circ}} \: \frac{q_1q_2}{ {d}^{3} } \vec{d}$

$\bf \: F = \frac{q_1q_2}{4\pi \epsilon_{ \circ}} \: \frac{(2 \hat{i} - \hat{j} - 3 \hat{k})}{14 \sqrt{14} }$

$\purple{ \boxed{ \bf{F = \frac{q_1q_2}{56 \sqrt{14} \pi \epsilon_{ \circ}} } \: (2 \hat{i} - \hat{j} + 3 \hat{k})}}$

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