Question

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Determine force on 2C charge in the following arrangement of charges kept at corners of a square of Side 2 m. Where k is Coulomb's constant ​

Answer 1

Answer:

Given:

A set of charges on the vertices of a square has been provided. Edge of square = 2 m , and Columns constant is given as k

To find:

Force on the 2C Charge.

Concept :

Consider the forces on the 2C charge as vector quantities.

The force due to -2C charges are perpendicular to one another and pulling in nature. Their resultant will be along the Diagonal.

The force due to 8√2 C charge will be pushing in nature and along the same diagonal.

Diagram :

Look at the attached photo to understand better.

Calculation:

Force due to one -2C charge :

$F1 = \dfrac{k(2 \times 2)}{ {2}^{2} } = k$

Similarly , force due to the the other -2C charge

$F2 = \dfrac{k(2 \times 2)}{ {2}^{2} } = k$

Now these are perpendicular to one another :

Resultant be R:

$R = \sqrt{ {k}^{2} + {k}^{2} } = k \sqrt{2}$

Similarly force due to 8√2 C charge:

$F3 = \dfrac{k(8 \sqrt{2} \times 2)}{ {(2\sqrt{2})}^{2} } = 2 \sqrt{2} k$

Now , F3 and R are opposite to one another :

So net force will be denoted Rnet.

$R_{net} = 2 \sqrt{2}k - \sqrt{2} k$

$R_{net} = \sqrt{2} k \:$

So final answer :

$\: \: \: \: \: \boxed{ \huge{ \sf{ \blue{R_{net} = \sqrt{2} k \:}}}}$

Answer 2

Answer:

$\large\boxed{\sf{1.2726\times{10}^{10}\;N}}$

Explanation:

There are two -2 C charges.

Let the magnitude of force on charge 2C due to -2C charges is $F_{1}$ and $F_{2}$ .

Distance of separation = 2 m

Therefore, by coulomb's law,

$F_{1}=F_{2}= k \dfrac{(2 \times 2)}{ {2}^{2} } = f$

where,

• $k = \dfrac{1}{4\pi {\epsilon}_{0}}=9\times{10}^{9} \;N{m}^{2}{C}^{-2}$

Now, both $F_{1}$ and $F_{2}$ are perpendicular to each other , therefore, their resultant will be equal to,

$= \sqrt{ {f}^{2} + {f}^{2} } = \sqrt{2 {f}^{2} } = \sqrt{2} f$

Also, there is a 8√2 C charge on the opposite vertex.

Let the magnitude of force is $F_{3}$

Distance of separation = 2√2 m ( diagonal length)

Therefore, by Coulomb's law,

$= > F_{3}= k \frac{(2 \times 8 \sqrt{2}) }{ {(2 \sqrt{2} ) }^{2} } \\ \\ = > F_{3}= k \frac{(2 \times 2)}{ {2}^{2} } \times \frac{1}{2} \times \frac{1}{2} \times 8 \sqrt{2} \\ \\ = > F_{3} =2 \sqrt{2} f$

Now, The resultant of all forces is equal to

$= 2 \sqrt{2} f - \sqrt{2} f \\ \\ = \sqrt{2} f \\ \\ = \sqrt{2} \times 9 \times {10}^{9} \times \frac{(2 \times 2)}{ {2}^{2} } \\ \\ = 9 \sqrt{2} \times {10}^{9} \\ \\ = 9 \times 1.414 \times {10}^{9} \\ \\ = 12.726 \times {10}^{9} \\ \\ = 1.2726 \times {10}^{10}$

Hence, resultant force is $1.2726\times{10}^{10}\;N$