Question

Two point charges +2C and +6C repel each other with a force of 12N. If a charge of -4C is given to each of these charges, the force acting between them now is:​

Answer 1

We have,

• Initial charge on both points: +2C & +6C
• Force between them: 12 N

We know that, electrostatic force between two points is;

= F = k × q1q2/r²

• Putting the values from the ques.

= 12 = k × +2×(+6)/r²

= r² = k × 12/12

= r² = k

Now, the new charge will be

q1 = +2 -4 = -2 C

q2 = +6 - 4 = +2 --------------- eq1

Thus, the new electrostatic force will be;

= F = k × q1q2/r²

= F = k × -2 × 2/k

• put r² = k from eq 1

= F = -4N

So, the new force will be -4 N.

Answer 2

Answer:

Two point charges:

• Q₁ = +2 C
• Q₂ = +6 C
• Force of repulsion (F) = 12 N
• Distance between them = r

$\longrightarrow\rm F= K\dfrac{Q_1 \times Q_2}{r^2}$

$\longrightarrow\rm F= \dfrac{1}{4\pi \varepsilon _{0}} \times \dfrac{Q_1 \times Q_2}{r^2}$

$\longrightarrow\rm 12= 9 \times {10}^{ 9} \times \dfrac{2 \times 6}{r^2}$

$\longrightarrow\rm 12= 9 \times {10}^{ 9} \times \dfrac{12}{r^2}$

$\longrightarrow\purple{\rm {r}^{2} = 9 \times {10}^{ 9} }$

According to given condition, if a charge -4 C is added to each of them then the new charges will be :

• Q'₁ = +2 C - 4 C = -2C
• Q'₂ = +6 C - 4 C = 2 C

And also we know that,

$\longrightarrow\rm F \propto Q_1Q_2$

Therefore, the new force will be :

$\longrightarrow\rm \dfrac{F'}{F} =\dfrac{ Q'_1Q'_2}{Q_1 Q_2}$

$\longrightarrow\rm \dfrac{F'}{12} =\dfrac{ - 2 \times 2}{2 \times 6}$

$\longrightarrow\rm \dfrac{F'}{12} =\dfrac{ - 2 }{ 6}$

$\longrightarrow\rm \dfrac{F'}{12} =\dfrac{ - 1}{ 3}$

$\longrightarrow\rm3F'= - 12$

$\longrightarrow\rm F'= \dfrac{ - 12}{3}$

$\longrightarrow\underline{\boxed{\orange{\bold{ F'= - 4 \: N}}}}$