Question

Find the electrostatic force between two protons placed in free space separated by a distance
20cms
Ans 5.75 X 10 DNI​

Answer 1

Answer:

$\boxed{\sf Electrostatic \ force \ between \ two \ protons = 5.76 \times 10^{-25} \ N}$

Given:

Distance between two protons = 20 cm = 0.2 m

Explanation:

As we know;

$\sf Charge \ on \ proton \: (q)= 1.6 \times 10^{-19} \ C$

So,

$\sf Electrostatic \: Force \: (F) = \frac{k {q}^{2} }{ {r}^{2} } \\ \\ \sf = \frac{9 \times {10}^{9} \times {(1.6 \times {10}^{ - 19} )}^{2} }{ {(0.2)}^{2} } \\ \\ \sf = \frac{9 \times {10}^{9} \times 2.56 \times {10}^{ - 38} }{0.04} \\ \\ \sf = \frac{9 \times {10}^{9} \times 2.56 \times {10}^{ - 38} }{4 \times {10}^{ - 2} } \\ \\ \sf = \frac{9 \times 2.56 \times {10}^{(9 - 38 + 4)} }{4} \\ \\ \sf = \frac{9 \times 2.56 \times {10}^{ - 25} }{4} \\ \\ \sf = \frac{9 \times 0.64 \times \cancel4 \times {10}^{ - 25} }{ \cancel4} \\ \\ \sf = 9 \times 0.64 \times {10}^{ - 25} \\ \\ \sf = 5.76 \times {10}^{ - 25} \: N$