Question

Numerical Problems:-
(1.) What is the net force and its direction that the
Charges at the vertices A and c of the right triangle
angle ABC exert on the charge in vertex B ?​

Answer 1

Given that,

Distance between A and B = 4 cm

Distance between B and C = 2 cm

Charge on A = 7 μC

Charge on B = 2 μC

Charge on C = -2 μC

We need to calculate the force on A due to charge B

Using formula of force

$F_{AB}=\dfrac{kQQ_{1}}{r^2}$

Put the value into the formula

$F_{AB}=\dfrac{9\times10^{9}\times2\times10^{-6}\times7\times10^{-6}}{(4\times10^{-2})^2}$

$F_{AB}=78.75\ N$

We need to calculate the force on C due to charge B

Using formula of force

$F_{CB}=\dfrac{kQQ_{2}}{r^2}$

Put the value into the formula

$F_{CB}=\dfrac{9\times10^{9}\times2\times10^{-6}\times(-2\times10^{-6})}{(2\times10^{-2})^2}$

$F_{CB}=-90\ N$

We need to calculate the net force

Using formula for net force

$F_{net}=\sqrt{F_{1}^2+F_{2}^2}$

Put the value into the formula

$F_{net}=\sqrt{(78.75)^2+(-90)^2}$

$F_{net}=119.5\ N$

We need to calculate the direction of force

Using formula of direction

$\theta=\tan^{-1}(\dfrac{|F_{CB}|}{|F_{AB}|})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{90}{78.75})$

$\theta=48.8^{\circ}$

Hence, The net force is 119.5 N

The direction of force is 48.8°

Answer 2

Given that,

Distance between A and B = 4 cm

Distance between B and C = 2 cm

Charge on A = 7 μC

Charge on B = 2 μC

Charge on C = -2 μC

We need to calculate the force on A due to charge B

Using formula of force

$F_{AB}=\dfrac{kQQ_{1}}{r^2}$

Put the value into the formula

$F_{AB}=\dfrac{9\times10^{9}\times2\times10^{-6}\times7\times10^{-6}}{(4\times10^{-2})^2}$

$F_{AB}=78.75\ N$

We need to calculate the force on C due to charge B

Using formula of force

$F_{CB}=\dfrac{kQQ_{2}}{r^2}$

Put the value into the formula

$F_{CB}=\dfrac{9\times10^{9}\times2\times10^{-6}\times(-2\times10^{-6})}{(2\times10^{-2})^2}$

$F_{CB}=-90\ N$

We need to calculate the net force

Using formula for net force

$F_{net}=\sqrt{F_{1}^2+F_{2}^2}$

Put the value into the formula

$F_{net}=\sqrt{(78.75)^2+(-90)^2}$

$F_{net}=119.5\ N$

We need to calculate the direction of force

Using formula of direction

$\theta=\tan^{-1}(\dfrac{|F_{CB}|}{|F_{AB}|})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{90}{78.75})$

$\theta=48.8^{\circ}$

Hence, The net force is 119.5 N

The direction of force is 48.8°

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