Question

Two point charges Qa=8c and Qb=5c are separated by the distance r=10cm what is the magnitude of the force

Answer 1

AnswEr :

From the Question,

• $\sf Charge_1 = 8 \ C$

• $\sf Charge_2 = 5 \ C$

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

Electrostatic Force between the charges would be given as :

$\huge{\boxed{\boxed{\sf F = \dfrac{1}{4 \pi \epsilon_o} \times \dfrac{Q_a Q_b}{r^2}}}}$

$\sf{Here}\begin{cases} \sf{Q_a,Q_b \longrightarrow Charges } \\ \sf{r \longrightarrow Distance \ Of \ Separation } \\ \sf{\epsilon_o \longrightarrow Permittivity \ Of \ Free \ Space} \end{cases}$

Putting the values,

$\longrightarrow \: \sf \: F = 9 \times {10}^{9} \times \dfrac{8 \times 5}{ ({ {10}^{ - 1} )}^{2} } \\ \\ \longrightarrow \: \sf \: F = 40 \times 9 \times {10}^{11} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf F = 36 \times {10}^{12} \: N}}$

The electrostatic force acting on the charges is of the magnitude 36 × 10¹² N

Answer 2

Answer:

The Magnitude of the force (F) is 3.6 × 10¹³ N

Given:

1. First Charge (Q_a) = 8 C

2. Second Charge(Q_b) = 5 C

3. Distance of separation (r) = 10 cm = 0.1 m

Explanation:

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From the formula we know,

⇒ F = ( 1 / 4 π ε₀ ) × ( {Q_a Q_b} / r² )

Substituting the values,

⇒ F = (1 / 4 π ε₀ ) × ( {8 × 5} / [0.1]² )

⇒ F = (1 / 4 π ε₀ ) × ( 40 / 0.01 )

⇒ F = (1 / 4 π ε₀ ) × 40 × 10²

⇒ F = 9 × 10⁹ × 40 × 10²

         ∵ [ (1 / 4 π ε₀ ) = 9 × 10⁹ ]

⇒ F = 360 × 10¹¹

⇒ F = 3.6 × 10¹³

⇒ F = 3.6 × 10¹³ N

∴ The Magnitude of the force (F) is 3.6 × 10¹³ N

Note:

• Symbols have their usual meanings.

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