Question

Two electrical point charges are attracting each other with a force of 10 N. If the distance between them is reduced by 40% then the force between them Select one: a. increases to 62.5 N b. increases to 40 N c. decreases to 6.25 N d. decreases to 2.5 N e. Not changed f. increases to 25 N

Answer 1

Answer:

27.78 N

Or, 62.5N

Explanation:

Electric force is directly proportional to product of magnitude of charge and inversely proportional to square of distance between them.

F = k $\bold{\dfrac{q_1 q_2}{r^2}}$

*where k be any constant.

In question -

= > 10N = k $\bold{\dfrac{q_1 q_2}{r^2}}$ ... (1)

When distance is reduced by 40%:

New distance = r - 40%r = r( 1 - 40% )

New distance = r( 1 - 40/100)

New distance = 3r/5

Therefore, new force -

= > k$\bold{\dfrac{q_1 q_2}{\bigg(\dfrac{3r}{5}\bigg)^2}}$

= > k$\bold{\dfrac{q_1 q_2}{\dfrac{9}{25}r^2}}$

= > k$\bold{\dfrac{25}{9}\times\dfrac{q_1 q_2}{r^2}}$

From (1),

= > $\dfrac{25}{9}\times10$ N

= > 27.77 N

If your question is "reduces to* 40%"

New distance = 40% of r = 0.4 r

New force = 10N/(0.4)² = 62.5N