Question

please answer plzplz​

Answer 1

Solution :

⏭ Given:

✏ Two charged particles q and 4q are kept at distance 3m apart.

✏ A charged particle of charge Q is placed at point P.

⏭ To Find:

✏ Net force on Q charged particle due to q and 4q

⏭ Concept:

✏ This question is completely based on concept of coulomb's law of force between charge particles.

✏ Since, all particles have positive charge...Force acting between them is repulsive type.

⏭ Formula:

✏ As per coulomb's law, force between two charged particle is given by

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{ \large{F = \frac{1}{4\pi \epsilon_ {o}} \times \frac{q _ 1q _ 2}{ {r}^{2} } }}}}}} \: \star$

⏭ Calculation:

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• Force between q and Q

$\mapsto \sf \: F_1 = \dfrac{1}{4\pi \epsilon_{o}} \times \dfrac{(q)(Q)}{ {1}^{2} } \\ \\ \mapsto \sf \: \red{F_1 = \dfrac{Qq}{4\pi \epsilon_{o}}}$

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• Force between 4q and Q

$\mapsto \sf \: F_2 = \dfrac{1}{4\pi \epsilon_{o}} \times \dfrac{(4q)(Q)}{ {2}^{2} } \\ \\ \mapsto \sf \: \blue{F_2 = \dfrac{Qq}{4\pi \epsilon_{o}} }$

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• Net force on Q

$\mapsto \sf \: \green{F_1 = F_2 = F} \\ \\ \mapsto \sf \: F_{net} = F_1 - F_2 \\ \\ \mapsto \sf \: F_{net} = F - F \\ \\ \mapsto \: \underline{ \boxed{ \bold{ \sf{ \orange{ \large{F_{net} = 0}}}}}} \: \gray{ \surd}$

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