Question

Two point charges of 3 and 4 microcoulomb repel each other with a force of 10 N. If each is given an additional charge of -6 microcoulomb.what is the new force?

Answer 1

The new force F2 will be 5 N

Given

Charge q1 = +3 μC

Charge q2 = +4 μC

Force F1 = 10 N

Additional charge given to each = -6 μC

To Find

New Force F2

Solution

We can solve the numerical by following the steps below -

It is given that the,

Initial charge q1 = +3 μC

and q2 = +4 μC

After giving the additional charge of -6 μC to each,

The final charge q1' = +3 + (-6) μC

q1' = -3 μC

Similarly final charge q2' = +4 + (-6)

q2' = -2 μC

Now, according to Coulomb's Law,

$F = \frac{kq1q2}{r^{2}}$

So Initial force, F1 is

$F1 = \frac{kq1q2}{r^{2} }$

$F1 = \frac{k*3*4}{r^{2} }$ = 10 N (given)

The new force F2 is

$F2 = \frac{kq1'q2'}{r^{2} }$

$F2 = \frac{k (-3)*(-2)}{r^{2} }$

Taking the ratio of the two forces we have,

$\frac{F1}{F2} = \frac{3*4}{(-3)*(-2)} \\\\\frac{10}{F2} = \frac{12}{6} \\\\F2 = \frac{10}{2}\\\\ F2 = 5 N$

Therefore, the new force F2 will be 5 N.

Answer 2

Answer:

The new force is 5N.

Explanation:

Given:

Let the initial charges be q1 and q2 where

q1=3μC and q2=4μC

F=10N

Final charges be q1' and q2'

where q1'=3μC-6μC= -3μC and

q2'=4μC-6μC= -2μC

According to coulombs law ,we know that

F=k(q1×q2)/$r^{2}$

⇒F ∝ q1×q2

∴F/F'=(q1×q2)/(q1'×q2')

10/F'=(3×4)/(-3×-2)

F'=5N

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