Question

5. Two electric charges one 15 times as strong as the other, exert a force 2.45×10-2 on each other. When they are placed 10 cm apart in air, find the magnitude of each charge.​

Answer 1

AnswEr :

From the Question,

• Electrostatic Force (F) = 2.45 × 10^{-2} N

• Distance of Separation (r) = 10 cm = 10^{-1} m

If one of the charge is Q,the other charge would be 15 Q

We know that,

$\sf \: F = \dfrac{1}{4 \pi \epsilon_o} \times \dfrac{Qq}{r {}^{2} }$

$\mathbb{HERE}\begin{cases}\sf{F \longrightarrow Force} \\ \sf{\epsilon_o \longrightarrow Permittivity \ of \ Free \ Space} \\ \sf{Q,q \longrightarrow Charges } \\ \sf{r \longrightarrow Separation \ Distance } \end{cases}$

Also,

$\sf \dfrac{1}{4 \pi \epsilon_o} = 9 \times 10^9$

Substituting the values,we get :

$\sf \longrightarrow \: \: 2.45 \times {10}^{ - 2} = \dfrac{9 \times 10 {}^{9} \times {15q} \times q }{(10 {}^{ - 1} ) {}^{2} } \\ \\ \longrightarrow \: \sf \: 2.45 \times {10}^{ - 2} = \dfrac{135 {q}^{2} \times {10}^{9} }{ {10}^{ - 2} } \\ \\ \longrightarrow \: \sf \: 245 \times {10}^{ - 4} \times {10}^{ - 2} = 135 {q}^{2} \times {10}^{9} \\ \\ \longrightarrow \: \sf \: 135 {q}^{2} = 245 \times {10}^{ - 6} \times {10}^{ - 9} \\ \\ \longrightarrow \sf \: 135 {q}^{2} = 245 \times {10}^{ - 15} \\ \\ \longrightarrow \: \sf \: q {}^{2} = 1.81 \times {10}^{ - 15} \\ \\ \longrightarrow \: \sf \: q {}^{2} = 18.1 \times {10}^{ - 16} \\ \\ \longrightarrow \: \sf \: q = \sqrt{18.1 \times {10}^{ - 16} } \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: q = 4.25 \times {10}^{ - 8} \: C}}$

According to the Question,

The other charge should be 15Q

Thus,

$\sf \: Q = 15q \\ \\ \implies \: \sf \: Q = 15 \times 4.25 \times {10}^{ - 8} \\ \\ \implies \boxed{ \boxed{\sf \: Q = 63.75 \times {10}^{ - 8} \: C}}$

The charges are 63.75 × 10^{-8} C and 4.25 × 10^{-8} C