Question

A proton and an electron are placed 1.6 cm apart in free space. Find the magnitude of electrostatic force between them. What is the nature of this force.

Select one:
a. 90 x 10-25 repulsion
b. 9 x 10-25 repulsion
c. 9 x 10-25 N ; attractive
d. 9 x 1025 attractive​

Answer 1

Given :

• Distance (r) =1.6cm

To Find :

• The magnitude of electrostatics force between them =?

• The nature of this force =?

Formula :

• The magnitude of electrostatics force between them is

$\\ \red\bigstar\boxed{\sf{F=\dfrac{K q_{1} q_{2}}{r^{2}} }}$

Solution :

$\\ \bullet{\sf{Distance=1.6cm=16 \times 10^{-3} }}$

By using Formula :

$\\ \\ \implies\sf{F=\dfrac{K q_{1} q_{2} }{r^{2}}}$

$\\ \\ \\ \implies \sf \: force = \frac{9 \times {10}^{9} \times 1.6 \times {10}^{ - 19} \times 1.6 \times {10}^{ - 19} }{ {(16 \times {10}^{ - 3} })^{2} } \\ \\ \\ \\ \implies \sf \: force = \frac{9 \times {10}^{9} \times 16 \times {10}^{ - 20} \times 16 \times {10}^{ - 20} }{16 \times {10}^{ - 3} \times 16 \times {10}^{ - 3} } \\ \\ \\ \\ \implies \sf \: force = 9 \times {10}^{9} \times {10}^{ - 40} \times {10}^{6} \\ \\ \\ \\ \implies \sf \: force = 9 \times {10}^{ - 25} \: newton$

$\\ \\ \\ \implies\boxed{\sf{Force(F)=9 \times 10^{-25} N}}$

• The magnitude of electrostatics force between them is $\sf{9 \times 10^{-25} N}$

>> Hence, the nature of force is Attractive .