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)

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

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