Question

यदि दो आवेशित कणो के बीच की दूरी 20% कम कर दी जाए तो बल कितना होगा?

Answer 1

Answer:

Force between two charges will increase by 56.25%

Explanation:

As we know that the electrostatic force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

now if the distance between them is reduced by 20%

then we have

$F' = \frac{kq_1q_2}{(0.80r)^2}$

$F' = 1.5625\frac{kq_1q_2}{r^2}$

$F' = 1.5625 F$

now percentage increase in the force is given as

$percentage = \frac{F' - F}{F} \times 100$

$percentage = 56.25$%

#Learn

Topic : Electrostatic force

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