Question

Two charges of 2 and 8 µC are separated by 4 cm. Calculate the electrostatic force between them. If the distance between the charges is halved and a medium of dielectric constant 2 is placed between them, find the new electrostatic force. Also find the change in force.
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Answer 1

Answer:

Refer above attachment.

Answer 2

Given:

• Two charges of 2uC and 8 uC are separated by 4 cm.

To find:

• Electrostatic force between the charges?

• Force when distance is halved and a dielectric constant 2 is placed?

• Change in force ?

Calculation:

Initial force is :

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

$\implies F = \dfrac{(9 \times {10}^{9}) (2 \times 8 \times {10}^{ - 12} )}{ {(0.04)}^{2} }$

$\implies F = \dfrac{(9 \times {10}^{9}) (16 \times {10}^{ - 12} )}{ 16 \times {10}^{-4} }$

$\implies F = 9 \times {10}^{9} \times {10}^{ - 8}$

$\implies F = 90 N$

So, force is 90 N.

In 2nd case :

$F2 = \dfrac{k(q1 \times q2)}{K \times {r}^{2} }$

$\implies F2 = \dfrac{(9 \times {10}^{9}) (2 \times 8 \times {10}^{ - 12} )}{ 2 \times {(0.02)}^{2} }$

$\implies F2 = \dfrac{(9 \times {10}^{9}) (16\times {10}^{ - 12} )}{ 8 \times {10}^{-4} }$

$\implies F2 = 9 \times {10}^{9} \times 2\times {10}^{ - 8}$

$\implies F2 = 180 N$

So, change in force is :

$\Delta F = (180- 90)$

$\implies \Delta F = 90 N$

Hope It Helps.