A particle a Having a charge of 2-0 - 10-6 C and a mass of 100 g is placed at the bottom of a smooth inclined plane oif inclination 30- Where should another particle B- having same charge and mass- be placed on the incline so that it may remain in equilibrium-

Q_a = 2 × 10^ 5 C, M_B = 80 g Let the distance be R unit For equilibrium F_e = mg sinθ ⇒ = m × 9.8 ×1/2 R_2 = 7.34 × 10^ 2m ⇒ R = 2.70924 × 10^ 1m From the bot ...

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Standard XII

Physics

Coulomb's Law - Vector Form

A particle a ...

Question

A particle a Having a charge of $2.0 \times 10^{- 6} C$ and a mass of 100 g is placed at the bottom of a smooth inclined plane oif inclination $30^{\circ}$ Where should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium?

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Solution

$Q_{a} = 2 \times 10^{- 5} C, M_{B} = 80 g$
Let the distance be R unit
For equilibrium
$F_{e} = m g sin θ \Rightarrow= m \times 9.8 \times \frac{1}{2} R_{2} = 7.34 \times 10^{- 2} m \Rightarrow R = 2.70924 \times 10^{- 1} m$
From the bottom.

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A particle a Having a charge of 2.0×10−6C and a mass of 100 g is placed at the bottom of a smooth inclined plane oif inclination 30∘ Where should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium?

Qa=2×10−5C,MB=80g Let the distance be R unit For equilibrium Fe=mgsinθ⇒=m×9.8×12R2=7.34×10−2m⇒R=2.70924×10−1m From the bottom.

A particle a Having a charge of $2.0 \times 10^{- 6} C$ and a mass of 100 g is placed at the bottom of a smooth inclined plane oif inclination $30^{\circ}$ Where should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium?

$Q_{a} = 2 \times 10^{- 5} C, M_{B} = 80 g$
Let the distance be R unit
For equilibrium
$F_{e} = m g sin θ \Rightarrow= m \times 9.8 \times \frac{1}{2} R_{2} = 7.34 \times 10^{- 2} m \Rightarrow R = 2.70924 \times 10^{- 1} m$
From the bottom.