Question

$\large{\mathfrak{Bonjour!♡}}$•°⭑
Find the Fnet on charge at vertex B.
#refer to the above diagram
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Answer 1

Answer:

$\large\boxed{\sf{f_{net} = \dfrac{ \sqrt{3} }{4\pi{\epsilon}_{0}} \dfrac{ {q}^{2} }{ {a}^{2} } \;\;N}}$

Explanation:

It's being given that an equilateral triangle of side length 'a' is there.

There are three charges each of magnitude +q placed at the vertices .

Now, we have to find the net force on charge placed at vertex B.

Clearly, the charges at A and C will exert the repulsive forces on charge at B.

Also, distance of separation is 'a' for both the vertices.

Since, the charges are at equal distance and in equal magnitude,

Therefore, the force will also be same.

Let the magnitude of force be denoted by 'f'.

Therefore, we have,

By Coulomb's Law,

$= > f = \dfrac{1}{4\pi{\epsilon}_{0}} \dfrac{ {q}^{2} }{ {a}^{2} }$

Now, clearly, the forces are due to charges at A and C and at an angle of 60°.

$\red{Note:-}$ Refer to the attachment.

Now, for the resultant force, we have,

$= > f_{net} = \sqrt{ {f}^{2} + {f}^{2} + 2 {f}^{2} \cos(60) } \\ \\ = > f_{net} = \sqrt{ 2{f}^{2} + 2 {f}^{2} \times \frac{1}{2} } \\ \\ = > f_{net} = \sqrt{2 {f}^{2} + {f}^{2} } \\ \\ = > f_{net} = \sqrt{3 {f}^{2} } \\ \\ = > f_{net} = \sqrt{3} f \\ \\ = > f_{net} = \dfrac{ \sqrt{3} }{4\pi{\epsilon}_{0}} \dfrac{ {q}^{2} }{ {a}^{2} }$

Hence, the net force on charge at vertex B is $\bold{ \dfrac{ \sqrt{3} }{4\pi{\epsilon}_{0}} \dfrac{ {q}^{2} }{ {a}^{2} } \;N}$