Question

Two point charges 1C each separated by 1Km . Find the electrostatic force of

repulsion between them​

Answer 1

Solution :

⏭ Given:

✏ Two point charges of 1C are separated by 1km.

⏭ To Find:

✏ Electrostatic force of repulsion between them.

⏭ Formula:

✏ As per coulomb's law, Formula of electrostatic force between two point charges is given by...

$\star \: \underline{ \boxed{ \bold{ \pink{\sf{F = \dfrac{1}{4\pi \epsilon_o } \times \dfrac{q1q2}{ {r}^{2} } }}}}} \: \star$

⏭ Coversation:

✏ 1km = 1000m

⏭ Calculation:

$\mapsto \sf \: F = \dfrac{1}{4 \pi \epsilon _{o} } \times \dfrac{1 \times 1}{ { ({10}^{3}) }^{2} } \\ \\ \mapsto \sf \: f = \dfrac{9 \times {10}^{9} }{ {10}^{6} } \\ \\ \gray{ \bigstar} \: \underline{ \boxed{ \bold {\large{ \sf{ \purple{F = 9 \: kN}}}}}} \: \gray{ \bigstar}$

Answer 2

Given ,

• The magnitude of two charges is 1 C
• The distance b/w two charges is 1 km or (10)^3 m

We know that , the electrostatic force between two charges is given by

$\large \mathtt{\fbox{Electrostatic \: force = k\frac{ q_{1} q_{2} }{ {(r)}^{2} } }}$

Thus ,

$\mapsto \sf Electrostatic \: force = \frac{ 9 \times {(10)}^{9} \times {(1)}^{2} }{ {(1000) }^{2} } \\ \\ \mapsto \sf Electrostatic \: force =9 \times {(10)}^{(9 - 6)} \\ \\ \mapsto \sf Electrostatic \: force =9 \times {(10)}^{3} \: \: \: newton$

Hence , the electrostatic repulsive force is 9 × (10)^3 newton

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