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2 years ago

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Object A, which has been charged to +13.5 nC, is at the origin. Object B, which has been charged to -22.05 nC, is at x=0 and y=1

.84 cm. What is the magnitude of the electric force on object A?

1 answer:

bezimeni [28]2 years ago

6 0

Answer:

7.9 x 10^-3 N attractive in nature

Explanation:

QA = 13.5 nC = 13.5 x 10^-9 C

QB = - 22.05 nC = - 22.05 x 10^-9 C

d = 1.84 cm = 0.0184 m

Let the electric force on object A is F.

By use of Coulomb's law in electrostatics

F = K x QA x QB / d^2

F = (9 x 10^9 x 13.5 x 10^-9 x 22.05 x 10^-9) / (0.0184 x 0.0184)

F = 7.9 x 10^-3 N attractive in nature

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Case B :

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Case C :

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Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

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Case D :

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2 years ago

Read 2 more answers

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Answer:

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$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

=> $\Delta KE = 255.38m$

$255.38m = m * 9.8 * \mu_k * 36.0$

=> $255.38 = 352.8 * \mu_k$

=> $\mu_k = 0.724$

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2 years ago

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Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

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Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

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$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0

2 years ago