Question

calculate electrostatic godse between two protons which are 10*-15m part​

Answer 1

Solution

The repulsive forces are of magnitude 237.6 N

From the Question,

• Charge on a porton is 1.6 × 10^-19 C

• Distance of seperation,d = 10^-15 m

To finD

The electrostatic force between the two protons

Using the Relation,

$\huge{ \boxed{ \boxed{ \sf{F = \dfrac{KQq}{ {d}^{2} } }}}}$

Putting the values,we get :

$\colon \implies \: \sf \: F = 9 \times {10}^{9} \times \dfrac{(1.6 \times {10}^{ - 19}) {}^{2} }{ { ({10}^{- 15} )}^{2} } \\ \\ \colon \implies \: \sf \: F = 9 \times {10}^{9} \times 2.64 \times {10}^{ - 8} \\ \\ \colon \implies \: \boxed{ \boxed{ \sf{F = 237.6 N}}}$

Answer 2

≡QUESTION≡

Calculate electrostatic force between two protons which are 10⁻¹⁵m apart​.

                                                   

║⊕ANSWER⊕║

Charge on a proton = 1.6 * 10⁻¹⁹ C

Distance of separation = 10⁻¹⁵ m

The Electrostatic force is given by the relation:

$F = \frac{KQq}{d^{2} }$

where k = 9 * 10⁹ N m²/ C²

$F = 9 * 10^{9} * \frac{(1.6 * 10^{-19)^2} }{(10^{-15} )^2}$

$F = 9 * 10^{9} * 2.64 * 10^{-8}$

F = 273.69 N