Question

Can you help me please

Answer 1

Answer :-

Given :-

• $\sf q_1 = -6 \times 10^{-6} C$
• $\sf F = 65 N$
• $\sf r = 0.05 m$

To Find :-

• Magnitude of second charge

Solution :-

We know that,

$\boxed{\sf F = \frac{Kq_1q_2}{r^2}}$

Substituting the values -

➩ $\sf 65 = \frac{9 \times 10^9 \times -6 \times 10^{-6} \times q_2}{(0.05)^2}$

➩ $\sf 65 = \frac{9 \times 6 \times 10^9 \times 10^{-6} \times q_2}{0.0025}$

➩ $\sf 65 = \frac{54 \times 10^{3} \times q_2}{25 \times 10^{-4}}$

➩ $\sf 65 = \frac{54\times 10^3 \times 10^{4}\times q_2}{25}$

➩ $\sf 65 \times 25 = 54 \times 10^7 q_2$

➩ $\sf q_2 = \frac{1625}{54\times 10^7}$

➩ $\boxed{\sf q_2 = 30 \times 10^{-7}}$