Question

Two charged boxes are 4m apart from each other the blue box has a charge of 0.000337 and is attracting the red box with a force of 626N determine charge of the red box

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\large \sf Given \begin{cases} \tt{Distance \: (d) \: = \: 4 \: m} \\ \tt{Q_1 \: = \: 0.000337 \: C} \\ \tt{Force \: (F) \: = \: 626 \: N} \\ \tt{Q_2 \: = \: ?}\end{cases}$

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Solution :

$\sf{Q_1 \: = \: 0.000337 \: = \: 337 \: \times \: 10^{-6} \: C}$

Now, use Coloumb's Law :

$\huge{\boxed{\boxed{\rm{\blue{F \: = \: k \dfrac{Q_1 Q_2}{d^2}}}}}} \\ \\ \implies {\sf{Q_2 \: = \: \dfrac{F \: \times \: d^2}{Q_1 \: \times \: k}}} \\ \\ \implies {\sf{Q_2 \: = \: \dfrac{626 \: \times \: 4^2}{337 \: \times \: 10^{-6} \: \times \: 9 \: \times \: 10^{9}}}} \\ \\ \implies {\sf{Q_2 \: = \: \dfrac{626 \: \times \: 16}{337 \: \times \: 10^{-6} \: \times \: 9 \: \times \: 10^9}}} \\ \\ \implies {\sf{Q_2 \: = \: \dfrac{10016}{3033 \: \times \: 10^{3}}}} \\ \\ \implies {\sf{Q_2 \: = \: \dfrac{3.3}{10^{3}}}} \\ \\ \implies {\sf{Q_2 \: = \: 3.3 \: \times \: 10^{-3}}} \\ \\ {\underline{\sf{\therefore \: \: Charge \: of \: Red \: box \: is \: 3.3 \: \times \: 10^{-3} \: C}}}$

Answer 2

Answer:

Given:

Charge of blue box = 0.000337 C

Net Force = 626 N

Distance = 4 metres.

To find:

Charge on red box.

Concept:

Simple apply Coloumb's Law of Electrostatic Force :

$\: \: \: \: \: \: \: \: \: \boxed{ \red{ Force = \dfrac{1}{4\pi \epsilon_{o}} ( \dfrac{q_{1}q_{1}}{ {r}^{2} } ) }}$

Calculation:

Predetermined Value of :

$\: \: \: \: \: \: \: \boxed{ \blue {\dfrac{1}{4\pi \epsilon_{o}} \approx \: 9 \times {10}^{9} n {m}^{2} {c}^{ - 2}}}$

Let the charge of red box be q red.

Now putting all the available values given in the question :

$\rightarrow626 = 9 \times {10}^{9} \times ( \frac{3.37 \times {10}^{ - 4} \times q_{red}}{ {4}^{2} })$

$\rightarrow q_{red} = \dfrac{626 \times 16}{9 \times {10}^{9} \times 3.37 \times {10}^{ - 4} }$

$\rightarrow q_{red} \: = \: 330.23 \times {10}^{ - 5}C$

So final answer :

$\boxed{ \red{ \sf{ q_{red} \: = \: 330.23 \times {10}^{ - 5}C }}}$