Question

Two point charges + 6 μC and +16 μC repel each other with a force of 120 N. If a charge of -8 μC is added to each of them, then force bet ween them will becomes

40 N

30 N

20 N

60 N​

Answer 1

Answer:

20 N

Explanation:

Initially, charges are 6μC and 16μC. Later on, -8μC is added to both. Then, charges are (6 - 8)μC = -2μC and (16 - 8)μC = 8μC.

q = 6μC, Q = 16μC

q' = -2μC, Q' = 8μC

Here, everything remains(constant) same other-than charges, so let it A.

=> F = A(qQ)

=> 120 = A(6μC * 16μC) ...(1)

When -8μC is added, force is:

=> F' = A(q' Q')

=> F' = A(-2μC * 8μC) ...(2)

Divide (2) by (1):

=> F'/120 = A(-2μC * 8μC) / A(6μC * 16μC)

=> F'/120 = - 1/6

=> F' = - 120/6 N

=> F' = - 20 N

-ve sign just tells that it is opposite of the initial force. New force is 20N.

Answer 2

Answer:

If a charge of -8 μC is added to +6 μC and +16 μC, then the resultant force between them will become 20N and the force will act in the opposite direction to the initial force, i.e., the force will be attractive in nature.

Step by Step Explanation:

We can solve this question following the Coulomb's law equation:

$F=\frac{q_{1} q_{2} }{r^{2} }$

Calculation:

The force between the two-point charges initially:

$F_{i} =\frac{(+6)* (+16) }{r^{2} }$

⇒$F=\frac{96 }{r^{2} }$..............(1)

Now, a charge of -8 μC is added to both the charges.

∴ $q_{3} = (+6) + (-8) = -2$ μC.

and, $q_{4} = (+16) + (-8) = +8$ μC

∴ The new force between the two particles now is:

$F_{'} =\frac{(-2)* (+8) }{r^{2} }$

⇒ $F_{'} =\frac{-16 }{r^{2} }$.................(2)

Now, dividing eq. (2) by eq. (1):

$\frac{F_{'} }{F_{i} } = \frac{\frac{-16}{r^{2} } }{\frac{96}{r^{2} } }$

⇒ $\frac{F_{'} }{F_{i} } =\frac{-16}{96}$

⇒ $\frac{F_{'} }{120 } =\frac{-16}{96}$ [Initial force= 120N (given)]

⇒ $F_{'} = -20N$

Conclusion:

The final force between the two particles after the addition of -8 μC to both the charges is -20N.

The negative sign just indicates that the resultant force is acting in just the opposite direction as that of the initial force.

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