Q

# Confused! kindly explain, - Electrostatics - JEE Main-5

Two charges of $40\mu C$ and $-20\mu C$ are placed at a certain distance apart. They are touched and kept at the same distance. The ratio of the initial to the final force between them is

• Option 1)

8 : 1

• Option 2)

4 : 1

• Option 3)

1 : 8

• Option 4)

1 : 1

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As we learned @ 3224

Since only magnitude of charges are changes that’s why

$F\alpha\, q_{1}q_{2}\Rightarrow \frac{F_{1}}{F_{2}}=\frac{q_{1}q_{2}}{q'_{1}q'_{2}}=\frac{40\times 20}{10\times 10}=\frac{8}{1}$

Option 1)

8 : 1

Option 2)

4 : 1

Option 3)

1 : 8

Option 4)

1 : 1

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