Question

There are 2point charges +1ucolomb and -2ucoloumb placed at some distance.the ration of magnitude of electrostatic force acting on them is respectively???
1:2
2:1
1:4
1:1

Answer 1

Answer:

Given:

Point charges of +1 μC and -2 μC have been placed at some distance

To find:

Ratio of magnitude of Electrostatic force acting on them .

Concept:

As per the concept of Electrostatic force ;

It is

• Directly proportional to product of charges
• Inversely proportional to square of distance of separation.

Calculation:

Force experienced be F

$F = \dfrac{1}{4\pi \epsilon} \bigg \{\dfrac{(q1)(q2)}{ {r}^{2} } \bigg \}$

Putting available values:

$F = \dfrac{1}{4\pi \epsilon} \bigg \{\dfrac{(1 \times {10}^{ - 6} )(2 \times {10}^{ - 6} )}{ {r}^{2} } \bigg \}$

$F = \dfrac{1}{4\pi \epsilon} \bigg \{\dfrac{(2 \times {10}^{ - 12} )}{ {r}^{2} } \bigg \}$

Equal amount of force is being experienced by both the charges . Hence the ratio of force experienced should be 1:1.

$\boxed{ \red{ \bold{ \huge{ratio = 1 \: : \: 1}}}}$

Answer 2

GiveN :

• 1st Charge (Q1) = 1μC
• 2nd Charge (Q2) = -2μC

To FinD :

• Ratio of Electrostatic force

SolutioN :

Use Coulomb's Law :

$\dashrightarrow \large {\boxed{\tt{F \: = \: \dfrac{1}{4 \pi \epsilon _o} \dfrac{Q_1 Q_2}{d^2}}}} \\ \\ \dashrightarrow \tt{F \: = \: 9 \: \times \: 10^9 \: \times \: \dfrac{10^{-6} \: \times \: -2 \: \times \: 10^{-6}}{d^2}} \\ \\ \dashrightarrow \tt{9 \: \times \: 10^9 \: \times \: \dfrac{-2 \: \times \: 10^{-12}}{d^2}} \\ \\ \dashrightarrow \tt{F \: = \: \dfrac{-18 \: \times \: 10^{-3}}{d^2}}$

• If the charges are same and and the distance between charges is equal.

Then, the magnitude of force will be same

So, Dividing Them :

$\dashrightarrow \tt{Ratio \: = \: \dfrac{\dfrac{-18 \: \times \: 10^{-3}}{d^2}}{\dfrac{-18 \: \times \: 10^{-3}}{d^2}}} \\ \\ \dashrightarrow \tt{Ratio \: = \: \dfrac{1}{1}} \\ \\ \dashrightarrow {\boxed{\boxed{\tt{Ratio \: = \: 1:1}}}}$

✯ Ratio is 1:1