Question

When two point charges are 2.0cm apart , each one experiences a 1N electric force due to the other charge. If they are moved to a new separation of 8.0cm, calculate the force on each of them.​

Answer 1

Answer:1/16N

Explanation:refer to attachment please.

Answer 2

Given:

• 2 charges separated by 2 cm experience force of 1 N.

To find:

• Force when they are separated by 8 cm?

Calculation:

In the 1st case :

$F = \dfrac{k(q1 \times q2)}{ {r}^{2} }$

• Here r = 2 cm = 0.02 metres.

$\implies F = \dfrac{k(q1 \times q2)}{ {(0.02)}^{2} }$

• The value of F = 1 N.

$\implies F = \dfrac{k(q1 \times q2) \times {10}^{4} }{ {(2)}^{2} }$

$\implies F = \dfrac{k(q1 \times q2) \times {10}^{4} }{4 }$

$\implies \dfrac{k(q1 \times q2) \times {10}^{4} }{ {(2)}^{2} } = 1N$

In the final case :

• When r = 8 cm = 0.08 metres.

$\implies F2 = \dfrac{k(q1 \times q2)}{ {(0.08)}^{2} }$

$\implies F2 = \dfrac{k(q1 \times q2) \times {10}^{4} }{ {(8)}^{2} }$

$\implies F2 = \dfrac{k(q1 \times q2) \times {10}^{4} }{ 64}$

$\implies F2 = \dfrac{k(q1 \times q2) \times {10}^{4} }{16 \times 4}$

$\implies F2 = \dfrac{1}{16} \times \dfrac{k(q1 \times q2) \times {10}^{4} }{ 4}$

$\implies F2 = \dfrac{1}{16} \times F$

$\implies F2 = \dfrac{1}{16} \times 1$

$\implies F2 =0.0625 \: N$

So, final force is 0.0625 N.