Question

Twl point charges 3 micro coulomb and 8 micro coulomb repel each other with force 40 newton if charge of negative 5 is given to each of them then what will be the new force

Answer 1

Given that q1 = +2μC

q2 = +6μC

after -4μC is added, new charges

q3 = (+2-4)μC = -2μC

q4 = (+6-4)μC = +2μC

according to coloumb’s law of electrostatic force

F (force) between q1&q2 = Kq1q2/r2

F = (K/r2)* q1q2 = (K/r2) *2*6*10-12………..(since 1μC = 1*10-6C)….eqn1

F = (K/r2)*12*10-12

For q3 &q4

F/ = (K/r2)* -4*10-12………………..(here force is attractive force)…..eqn2

Taking ratio of eqn1 and eqn 2 we get,

F/F/ = 12/(-4)

Or F/ = -F/3

According to the problem, F = 12N hence F/ will be = -12/3 = -4N

Or attractive force of 4N

Hope it will help you

Answer 2

Answer:

The new force will be -10N

Explanation:

As the two point chargers repel each other.

The new charges will be

q¹=3-5=-2μC

q²=8-5=3μC

For q¹ it is given that F=40N

∴ 40=3*8/r²

For q² let the force be F

∴ F=-2*3/r²

∴F/40=-2*3/3*8=-10

Therefore, the new force will be -10N

Since force is a vector quantity the negative sign shows direction.