Question

Calculate force of repulsion between two electrons present at a distance of 0.001 m . ( use k = 9×10^9 Nm^2/C^2 )​

Answer 1

ANSWER

FOR POINTS CHARGES OR WE CAN SAY ELECTRON AND PROTON YOU CAN USE COLOUMBS LAW

JUST PUT VALUES YOU CAN GET YOUR ANSWER

BECAUSE I HAVE NO MACHINE WHICH PRODUCE FORCE BY PUTTING INPUT OF RADIUS AND CHARGE

HOPE IT HELPS

FOLLOW ME

BYE

Answer 2

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Electrostatics.

Now here, we know that, the force between the two electrons according to the Coulumbs law is given as,

Fe = ( K Q1.Q2) / r^2

Now for the force of repulsion we calculate as,

F = [(9*10^9) ( 1.6 * 10^-19)^2 ] ÷ (0.001)^2

so after solving we get as,

⭐F = 23.04 *10^-23 C^2/m^2

Since, here the distance is less, the force of repulsion is more.

as we can clearly see the relations as

⭐F is inversely proportional to the square of the distance between the charges.