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  • Can someone explain If a charge of 1 micro coulomb is divided into 2 parts in ratio 2:3 and placed 1 m apart in the vacuum. What will be the electric forces acting between in Newton?

If a charge of 1 micro coulomb is divided into 2 parts in ratio 2:3 and placed 1 m apart in the vacuum. What will be the electric forces acting between in Newton?

  • Option 1)

    0.00216

  • Option 2)

    216.0

  • Option 3)

    0.216

  • Option 4)

    0.0216

 

Answers (1)

best_answer

As we learnt in 

Coulombic force -

F\propto Q_{1}Q_{2}=F\propto \frac{Q_{1}Q_{2}}{r^{2}}=F=\frac{KQ_{1}Q_{2}}{r^{2}}

- wherein

K - proportionality Constant 

Q1 and Q2 are two Point charge

 

 Charges are q_{1} = \frac{2}{5}\mu C        and \: \: \: \: \ q_{2} = \frac{3}{5}\mu C

F = \frac{Kq_{1}q_{2}}{r^{2}} = \frac{9\times 10^{9}\times \frac{2}{5}\times \frac{3}{5}\times 10^{-12}}{1} = \frac{54}{25}\times 10^{-3}N

     = 2.16 \times 10^{-3}\ N = 0.00216\ N


Option 1)

0.00216

This option is correct

Option 2)

216.0

This option is incorrect

Option 3)

0.216

This option is incorrect

Option 4)

0.0216

This option is incorrect

Posted by

Plabita

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