Two charged particles each of mass 5g and charge q-2mu C are suspended as shown in the figure- The system is taken in a satellite- The force between the charges is-23 times 10-3- N2-3 times 10-3- N0-23 times 10-3- Nnone of these

Question

Two charged particles each of mass 5g and charge q-2mu C  are suspended as shown in the figure- The system is taken in a satellite- The force between the charges is-23 times 10-3- N2-3 times 10-3- N0-23 times 10-3- Nnone of these

Toppr Admin · Accepted Answer

Given - -xA0- l-60 cm -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- q-2-xD7-10-x2212-6 CAs there is no gravity- thus total distance between the charges -xA0- -xA0- d-2l-5-125 cm -1-25 mCoulomb force -xA0- -xA0- -xA0- -xA0-F-Kq2d2-9-xD7-109-xD7-2-xD7-10-x2212-6-2-1-25-2-23-xD7-10-x2212-3 N

Two charged particles each of mass $5$ g and charge $q = 2 μ C$ are suspended as shown in the figure. The system is taken in a satellite. The force between the charges is:

$23 \times 10^{- 3} N$
$2.3 \times 10^{- 3} N$
$0.23 \times 10^{- 3} N$
none of these

Given : $l = 60$ cm $q = 2 \times 10^{- 6}$ C
As there is no gravity, thus total distance between the charges $d = 2 l + 5 = 125$ cm $= 1.25$ m
Coulomb force $F = \frac{K q^{2}}{d^{2}} = \frac{9 \times 10^{9} \times (2 \times 10^{- 6})^{2}}{(1.25)^{2}} = 23 \times 10^{- 3}$ N

Two charged particles each of mass 5g and charge q=2μC are suspended as shown in the figure. The system is taken in a satellite. The force between the charges is:23×10−3N2.3×10−3N0.23×10−3Nnone of these

Given : l=60 cm q=2×10−6 CAs there is no gravity, thus total distance between the charges d=2l+5=125 cm =1.25 mCoulomb force F=Kq2d2=9×109×(2×10−6)2(1.25)2=23×10−3 N

Two charged particles each of mass $5$ g and charge $q = 2 μ C$ are suspended as shown in the figure. The system is taken in a satellite. The force between the charges is:

$23 \times 10^{- 3} N$
$2.3 \times 10^{- 3} N$
$0.23 \times 10^{- 3} N$
none of these

Given : $l = 60$ cm $q = 2 \times 10^{- 6}$ C
As there is no gravity, thus total distance between the charges $d = 2 l + 5 = 125$ cm $= 1.25$ m
Coulomb force $F = \frac{K q^{2}}{d^{2}} = \frac{9 \times 10^{9} \times (2 \times 10^{- 6})^{2}}{(1.25)^{2}} = 23 \times 10^{- 3}$ N