The electrostatic force between charges of 200 μC and 500 μC placed in free space is 5 gf. Find the distance between the two charges. Take g = 10 ms^–2.

Solution

charge - q_{1} = 200 \times 10^{- 6} C = 2 \times 10^{- 4} C, charge - q_{2} = 500 \times 10^{- 6} C = 5 \times 10^{- 4} C, Electrostatic force - F = 5 gf = 5 \times 10^{- 3} kgf = 5 \times 10^{- 3} \times 10 N = 5 \times 10^{- 2} N, w e h a v e t o f i n d t h e d i s \tan c e b e t w e e n t w o c h a r g e s i . e . r ?

Using the formula,

F = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r^{2}}, we get

5 \times 10^{- 2} = \frac{9 \times 10^{9} \times 2 \times 10^{- 4} \times 5 \times 10^{- 4}}{r^{2}}

\Rightarrow

r = 1.34 \times 10^{2} m \Rightarrow

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