Ncert Exercises & Solutions

A paisa coin is electrically neutral and contains equal amounts of positive and negative charge of magnitude \(34.8~\text{kC}\)34.8 kC. Suppose that these equal charges were concentrated in two-point charges separated by 1 cm, 100 m and 10⁶ m . Find the force on each such point charge in each of the three cases respectively.
1. \(1.09\times10^{23}~\text{N},~1.09\times10^{15}~\text{N}~\text{and}~1.09\times10^{7}~\text{N}\)
2. \(1.09\times10^{22}~\text{N},~1.09\times10^{15}~\text{N}~\text{and}~1.09\times10^{7}~\text{N}\)
3. \(1.09\times10^{23}~\text{N},~1.09\times10^{14}~\text{N}~\text{and}~1.09\times10^{8}~\text{N}\)
4. \(1.09\times10^{22}~\text{N},~1.09\times10^{15}~\text{N}~\text{and}~1.09\times10^{7}~\text{N}\)

Hint: Use the coulomb's law.

Step 1: Find the force between the charges in each case.

Here,

q = \pm 34.8 k C = \pm 3.48 \times 10^{4} C

r_{1} = 1 cm = 10^{- 2} m, r_{2} = 100 m, r_{3} = 10^{6} m and \frac{1}{4 π ε_{0}} = 9 \times 10^{9}

F_{1} = \frac{| q |^{2}}{4 π ε_{0} r_{1}^{2}} = \frac{9 \times 10^{9} \times {(3.48 \times 10^{4})}^{2}}{{(10^{- 2})}^{2}} = 1.09 \times 10^{23} N

F_{2} = \frac{| q |^{2}}{4 π ε_{0} r_{2}^{2}} = \frac{9 \times 10^{9} \times {(3.48 \times 10^{4})}^{2}}{(100)^{2}} = 1.09 \times 10^{15} N

F_{3} = \frac{| q |^{2}}{4 π ε_{0} r_{3}^{2}} = \frac{9 \times 10^{9} \times {(348 \times 10^{4})}^{2}}{{(10^{6})}^{2}} = 1.09 \times 10^{7} N

Step 2: Find the conclusion by observing the force between the charges.

When charges in an electrically neutral matter are separated by a distance, they exert a large force on each other and hence, it is no possible to separate the charges in an electrically neutral matter.

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